A106262 An invertible triangle of remainders of 2^n.
1, 0, 1, 0, 2, 1, 0, 1, 2, 1, 0, 2, 0, 2, 1, 0, 1, 0, 4, 2, 1, 0, 2, 0, 3, 4, 2, 1, 0, 1, 0, 1, 2, 4, 2, 1, 0, 2, 0, 2, 4, 1, 4, 2, 1, 0, 1, 0, 4, 2, 2, 0, 4, 2, 1, 0, 2, 0, 3, 4, 4, 0, 8, 4, 2, 1, 0, 1, 0, 1, 2, 1, 0, 7, 8, 4, 2, 1, 0, 2, 0, 2, 4, 2, 0, 5, 6, 8, 4, 2, 1, 0, 1, 0, 4, 2, 4, 0, 1, 2, 5, 8, 4, 2, 1
Offset: 0
Examples
Triangle begins: 1; 0, 1; 0, 2, 1; 0, 1, 2, 1; 0, 2, 0, 2, 1; 0, 1, 0, 4, 2, 1; 0, 2, 0, 3, 4, 2, 1; 0, 1, 0, 1, 2, 4, 2, 1; 0, 2, 0, 2, 4, 1, 4, 2, 1; 0, 1, 0, 4, 2, 2, 0, 4, 2, 1;
Links
- G. C. Greubel, Rows n = 0..50 of the triangle, flattened
Crossrefs
Programs
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Magma
[Modexp(2, n-k, k+2): k in [0..n], n in [0..15]]; // G. C. Greubel, Jan 10 2023
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Mathematica
Table[PowerMod[2, n-k, k+2], {n,0,15}, {k,0,n}]//Flatten (* G. C. Greubel, Jan 10 2023 *) -
SageMath
flatten([[power_mod(2,n-k,k+2) for k in range(n+1)] for n in range(16)]) # G. C. Greubel, Jan 10 2023
Formula
T(n, k) = 2^(n-k) mod (k+2).
Sum_{k=0..n} T(n, k) = A106263(n) (row sums).
Sum_{k=0..floor(n/2)} T(n-k, k) = A106264(n) (diagonal sums).
From G. C. Greubel, Jan 10 2023: (Start)
T(n, 0) = A000007(n).
T(n, 1) = A000034(n+1).
T(2*n, n) = A213859(n).
T(2*n, n-1) = A015910(n+1).
T(2*n, n+1) = A294390(n+3).
T(2*n+1, n-1) = A112983(n+1).
T(2*n+1, n+1) = A294389(n+3).
T(2*n-1, n-1) = A062173(n+1). (End)