A372523 Triangle read by rows: T(n, k) is equal to n/k if k | n, else to the concatenation of A003988(n, k) = floor(n/k) and A051127(k, n) = n mod k.
1, 2, 1, 3, 11, 1, 4, 2, 11, 1, 5, 21, 12, 11, 1, 6, 3, 2, 12, 11, 1, 7, 31, 21, 13, 12, 11, 1, 8, 4, 22, 2, 13, 12, 11, 1, 9, 41, 3, 21, 14, 13, 12, 11, 1, 10, 5, 31, 22, 2, 14, 13, 12, 11, 1, 11, 51, 32, 23, 21, 15, 14, 13, 12, 11, 1, 12, 6, 4, 3, 22, 2, 15, 14, 13, 12, 11, 1
Offset: 1
Examples
The triangle begins: 1; 2, 1; 3, 11, 1; 4, 2, 11, 1; 5, 21, 12, 11, 1; 6, 3, 2, 12, 11, 1; 7, 31, 21, 13, 12, 11, 1; ...
Links
- Stefano Spezia, First 150 rows of the triangle, flattened
Crossrefs
Programs
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Mathematica
T[n_,k_]:=If[Divisible[n,k],n/k,FromDigits[Join[IntegerDigits[Floor[n/k]],IntegerDigits[Mod[n,k]]]]]; Table[T[n,k],{n,12},{k,n}]//Flatten (* or *) T[n_,k_]:=Floor[n/k]10^IntegerLength[Mod[n,k]]+Mod[n,k]; Table[T[n,k],{n,12},{k,n}]//Flatten (* or *) T[n_, k_]:=SeriesCoefficient[x^k(1+Sum[(i + 10^(1+Floor[Log10[Mod[n,k]]]))*x^i, {i, k-1}] - Sum[i*x^(k+i), {i, k-1}])/(1-x^k)^2, {x, 0, n}]; Table[T[n, k], {n, 12}, {k, n}]//Flatten
Formula
T(n, k) = floor(n/k)*10^(1+floor(log10(n mod k))) + (n mod k) if n is not divisible by k.
T(n, n) = 1.
T(n, 1) = n.
T(n, k) = 2*T(n-k, k) - T(n-2*k, k) for n >= 3*k.
T(n, k) = [x^n] x^k*(1 + (Sum_{i=1..k-1} (i + 10^(1+floor(log10(n mod k))))*x^i) - (Sum_{i=1..k-1} i*x^(k+i)))/(1 - x^k)^2.
Comments