A341091 Triangle read by rows: Coefficients for calculation of the sum of all the finite differences from order zero to order k. Sum_{n=0..k} T(n, k)*b(n) = b(0) + b(1) + ... + b(k) + (b(1) - b(0)) + ... + (b(k) - b(k-1)) + ((b(2) - b(1)) - (b(1) - b(0))) + ... .
1, 0, 2, 1, -1, 3, 0, 3, -3, 4, 1, -2, 7, -6, 5, 0, 4, -8, 14, -10, 6, 1, -3, 13, -21, 25, -15, 7, 0, 5, -15, 35, -45, 41, -21, 8, 1, -4, 21, -49, 81, -85, 63, -28, 9, 0, 6, -24, 71, -129, 167, -147, 92, -36, 10, 1, -5, 31, -94, 201, -295, 315, -238, 129, -45, 11
Offset: 0
Examples
Triangle begins with T(n, k): n= 0, 1, 2, 3, 4, 5, 6, 7, 8 k=0 1 k=1 0, 2 k=2 1, -1, 3 k=3 0, 3, -3, 4 k=4 1, -2, 7, -6, 5 k=5 0, 4, -8, 14, -10, 6 k=6 1, -3, 13, -21, 25, -15, 7 k=7 0, 5, -15, 35, -45, 41, -21, 8 k=8 1, -4, 21, -49, 81, -85, 63, -28, 9 ...
Crossrefs
Sequences below will be obtained by evaluation of the associated polynomials:
Programs
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PARI
A341091(n, k) = sum(m=n, k,(-1)^(m+n)*binomial(m+1, n))
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PARI
A341091(n, k) = (1/2)*(-1)^n*(2*(-1)^k*binomial(2+k, n)*hypergeom([1,k+3],k+3-n,-1)+(-1/2)^n*(2^(n+1)-1)) \\ Thomas Scheuerle, Apr 29 2024
Formula
b(0)*T(0, m) + b(1)*T(1, m) + ... + b(m)*T(m, m)
= Sum_{j=0..m} Sum_{n=0..m-j} Sum_{k=0..n} (-1)^k*binomial(n, k)*b(j+n-k)
= Sum_{n=0..m} b(n)*Sum_{j=n..m}(-1)^(j+n)*binomial(j+1, n).
T(n, k) = Sum_{m=n..k}(-1)^(m+n)*binomial(m+1, n).
T(n, k) = (1/2)*(-1)^n*(2*(-1)^k*binomial(2+k, n)*Hypergeometric2F1(1, k+3, k+3-n, -1)+(-1/2)^n*(2^(n+1) - 1)), where Hypergeometric2F1 is the Gaussian hypergeometric function 2F1 as defined in Mathematica. - Thomas Scheuerle, Apr 29 2024
T(k, k) = A000027(k+1) The positive integers.
|T(k-1, k)| = A000217(k) The triangular numbers.
T(k-2, k) = A004006(k).
|T(k-3, k)| = A051744(k).
T(0, k*2) = 1.
T(0, k*2 + 1) = 0.
T(1, k*2 + 1) = k + 2.
T(1, k*2 + 2) = -(k + 1).
T(n, k) with constant n and variable k, a linear recurrence relation with characteristic polynomial (x-1)*(x+1)^(n+1).
Sum_{n=0..k} T(n, k)*B_n = 1. B_n is the n-th Bernoulli number with B_1 = 1/2. B_n = A164555(n)/A027642(n).
Sum_{n=0..k} T(n, k)*(1 - B_n) = k.
Sum_{n=0..k} T(n, k)*(2*n - 3+3*B_n) = k^2.
From Thomas Scheuerle, Apr 29 2024: (Start)
Sum_{n=0..k} T(n, k)*(1/(1+n)) = H(1+floor(k/2)), where H(k) is the harmonic number A001008(k)/A002805(k). (End)
Sum_{n=0..k} T(n, k)*c(n) = c(k). C(k) = {-1, 0, 1/2, 1/2, 1/8, -7/20, ...} this sequence of rational numbers can be defined recursively: c(0) = -1, c(m) = (-c(m-1) + Sum_{k=0..m-1} A130595(m+1, k)*c(k))/m.
c(m) is an eigensequence of this transformation, all eigensequences are c(m) multiplied by any factor.
Sum_{n=0..k} T(n, k)*A000045(n) = 2*(A000045(2*floor((k+1)/2) - 1) - 1). A000045 are the Fibonacci numbers.
This sequence acting as an operator onto a monomial n^w:
Sum_{n=0..k} T(n, k)*n^w = (1/(w+1))*k^(w+1) + Sum_{v=1..w} ((v+B_v)*(w)_v/v!)*k^(w+1-v) - A052875(w) + O_k(w) (w)_v is the falling factorial. If k > w-1 then O_k(w) = 0. If k <= w-1 then O_k(w) is A084416(w, 2+k), the sequence with the exponential generating function: (e^x-1)^(2+k)/(2-e^x).
From Thomas Scheuerle, Apr 29 2024: (Start)
This sequence acting by its inverse operator onto a monomial k^w:
Sum_{n=0..k} T(n, k)*( Sum_{m=0..k} ((-1)^(1+m+k)*binomial(k, m)*(2^(k-m) - 1)*n^m + A344037(m)*B_n) ) = k^w - A372245(w, k+3), note that A372245(w, k+3) = 0 if k+3 > w. B_n is the n-th Bernoulli number with B_1 = 1/2.
How this sequence will act as an operator onto a Dirichlet series may be developed by the formulas below:
Sum_{n=0..k} T(n, k)*2^n = A000295(k+2).
Sum_{n=0..k} T(n, k)*3^n = A000392(k+3).
Sum_{n=0..k} T(n, k)*4^n = A016208(k).
Sum_{n=0..k} T(n, k)*5^n = A016218(k).
Sum_{n=0..k} T(n, k)*6^n = A016228(k).
Sum_{n=0..k} T(n, k)*7^n = A016241(k).
Sum_{n=0..k} T(n, k)*8^n = A016249(k).
Sum_{n=0..k} T(n, k)*9^n = A016256(k).
Sum_{n=0..k} T(n, k)*10^n = A016261(k).
Sum_{n=0..k} T(n, k)*m^n = m^2*m^k/(m-1) - (m-1)^2*(m-1)^k/(m-2) + 1/((m-1)*(m-2)), for m > 2.
Sum_{n=0..k} T(n, k)*( m*B_n + (m-1)*Sum_{t=1..m} t^n )*(1/m^2) = m^k, for m > 0. B_n is the n-th Bernoulli number with B_1 = 1/2.
Sum_{n=0..k} T(n, k) zeta(-n) = Sum_{j=0..k} (-1)^(1+j)/(2+j) = (-1)^(k+1)*LerchPhi(-1, 1, k+3) - 1 + log(2).
Sum_{n=0..k} T(k - n, k)*2^n = A000975(k+1)
Sum_{n=0..k} T(k - n, k)*3^n = A091002(k+2)
Sum_{n=0..k} T(k - n, k)*4^n = A249997(k). (End)
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