A128315 Inverse Moebius transform of signed A007318.
1, 0, 1, 2, -2, 1, -1, 4, -3, 1, 2, -4, 6, -4, 1, 0, 4, -9, 10, -5, 1, 2, -6, 15, -20, 15, -6, 1, -2, 11, -24, 36, -35, 21, -7, 1, 3, -10, 29, -56, 70, -56, 28, -8, 1, 0, 6, -30, 80, -125, 126, -84, 36, -9, 1, 2, -10, 45, -120, 210, -252, 210, -120, 45, -10, 1, -2, 18, -67, 176, -335, 463, -462, 330, -165, 55, -11, 1
Offset: 1
Examples
First few rows of the triangle: 1; 0, 1; 2, -2, 1; -1, 4, -3, 1; 2, -4, 6, -4, 1; 0, 4, -9, 10, -5, 1; ...
Links
- G. C. Greubel, Rows n = 1..100 of the triangle, flattened
Crossrefs
Programs
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Magma
A128315:= func< n,k | (&+[0^(n mod j)*(-1)^(k+j)*Binomial(j-1, k-1): j in [k..n]]) >; [A128315(n,k): k in [1..n], n in [1..15]]; // G. C. Greubel, Jun 22 2024
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Mathematica
A128315[n_, k_]:= (-1)^k*DivisorSum[n, (-1)^#*Binomial[#-1, k-1] &]; Table[A128315[n,k], {n,15}, {k,n}]//Flatten (* G. C. Greubel, Jun 22 2024 *)
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SageMath
def A128315(n,k): return sum( 0^(n%j)*(-1)^(k+j)*binomial(j-1,k-1) for j in range(k,n+1)) flatten([[A128315(n,k) for k in range(1,n+1)] for n in range(1,16)]) # G. C. Greubel, Jun 22 2024
Formula
T(n, 1) = A048272(n).
Sum_{k=1..n} T(n, k) = A000012(n) = 1 (row sums).
From G. C. Greubel, Jun 22 2024: (Start)
T(n, k) = (-1)^k * Sum_{d|n} (-1)^d * binomial(d-1, k-1).
T(n, 2) = A325940(n), n >= 2.
T(n, 3) = A363615(n), n >= 3.
T(n, 4) = A363616(n), n >= 4.
T(2*n-1, n) = (-1)^(n-1)*A000984(n-1), n >= 1.
Sum_{k=1..n} (-1)^(k-1)*T(n, k) = (-1)^(n-1)*A081295(n).
Sum_{k=1..n} k*T(n, k) = A000034(n-1), n >= 1.
Sum_{k=1..n} (k+1)*T(n, k) = A010693(n-1), n >= 1. (End)
Extensions
a(43) = 28 inserted and more terms from Georg Fischer, Jun 05 2023
Comments