Divisor-poset inversion of an affine function

Goal and scope

For f(k)=2k+1 at n=6, F(6)=28 and inversion returns f(6)=13; n=1 roundtrips too.

Every divisor of positive integer 6 and an exact affine family; only this finite inversion certificate is claimed.

Givens and symbols

• calc

• G1 — f(k)=2k+1, F(m)=sum of f(d) over d dividing m; n=6, with indexing mu(d)*F(6/d).

• F — divisor forward transform
• f — affine function 2k+1

Rules and sources

• calc · prerequisite — Assume exact integer arithmetic, division with remainder, and definitions of divisibility and gcd; every congruence has an explicit modulus.
• mko_definition · cited — For positive integer n, F(n)=sum over d|n of f(d) implies f(n)=sum over d|n of mu(d)*F(n/d). The equivalent indexing mu(n/d)*F(d) is valid but must not be mixed with the first convention. Enumerate every positive divisor and derive mu from complete prime multiplicities. Retain n=1 and mu(1)=1. This is convolution on the divisor poset, not a sum over all d≤n. · mko-mobius-inversion-finite-divisors

Main derivation

6=2\cdot3

D(6)=\{1,2,3,6\}

d=1,2,3,6

f(d)=3,5,7,13\text{ respectively}

D(1)=\{1\},\ D(2)=\{1,2\},\ D(3)=\{1,3\}

F(1)=3,\quad F(2)=3+5=8,\quad F(3)=3+7=10

D(6)=\{1,2,3,6\}

F(6)=3+5+7+13=28

d=1,2,3,6

\mu(d)=1,-1,-1,1;\quad6/d=6,3,2,1

\sum_{d\mid6}\mu(d)F(6/d)

28-10-8+3=13=f(6)

F(6)-F(3)-F(2)+F(1)

f(1)(1-1-1+1)+f(2)(1-1)+f(3)(1-1)+f(6)=f(6)

n=1,\quad D(1)=\{1\}

F(1)=f(1)=3,\quad\mu(1)F(1)=3

Substeps and lemmas

No substeps in this guide.

Reverse audits

• S08 · premise_audit · G1, S03, S05, S07 — Zero, negatives and incomplete divisor lists are outside this certificate.
• S07 · premise_audit · G1, S03, S04, S06 — Terms appear according to divisibility, not an ordinary d≤n prefix sum.
• S06 · premise_audit · G1, S03, S04, S05 — The fully reindexed mu(6/d)F(d) is also valid; changing only one side is not.
• S05 · premise_audit · G1, S01 — d=6 pairs with F(1), not F(6).
• S04 · premise_audit · G1, S01, S02 — Using only proper divisors gives 15, not 28, breaking the roundtrip.
• S03 · premise_audit · G1, S01, S02 — F(3) excludes f(2) because 2 does not divide 3.
• S02 · premise_audit · G1, S01 — Only these four f-values are used, not all integers 1 through 6.
• S01 · premise_audit · G1 — 4 and 5 are below 6 but do not divide it.

Error controls and negative controls

Index pairing

Each term is mu(d)F(n/d), or a fully reindexed mu(n/d)F(d).

All divisors

Include 1 and n, not just prime or proper divisors.

• F(6) includes f(4),f(5). → reject — 4 and 5 are not divisors of 6.
• Use F(d) with unchanged mu(d) in every inverse term. → reject — Here 3-8-10+28 happens to be 13, but at n=2,3-8=-5 differs from 5; correct reindexing changes mu to mu(n/d).
• Prime divisors alone suffice for inversion. → reject — Omitting 1 and 6 leaves -10-8=-18, not 13.

Result and limitations

For f(k)=2k+1 at n=6, F(6)=28 and inversion returns f(6)=13; n=1 roundtrips too.

Verification and source binding

2 declared checks for this guide; only the listed expressions are covered.

• source commit: b1b2f1dcf8c50d67a93fe52d95f8eef7b3abba9f
• adopted source SHA-256: a3d906c53553c720d08fa4763944293d886bfad9ea166a77f4ae2c1399c65f18
• original source SHA-256: a3d906c53553c720d08fa4763944293d886bfad9ea166a77f4ae2c1399c65f18
• Lean: NOT_RUN · independent AI: SCOPED_LEAD_AND_BILINGUAL_CONTEXT_REVIEW

Same-source full JSON (formulas, ASTs, controls and revision log)