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

1,2

Links

Stefano Spezia, First 150 rows of the triangle, flattened

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.

Example

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;
  ...

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

Crossrefs

Cf. A003988, A004216, A051127, A055642.

Cf. A000012 (right diagonal), A000027 (1st column).

Cf. A048158, A051126, A051127, A234575.

Sequence in context: A203709 A270382 A340063 * A196371 A276116 A306993

Adjacent sequences: A372520 A372521 A372522 * A372524 A372525 A372526

Keyword

nonn,base,easy,look,tabl

Author

Stefano Spezia, May 04 2024

Status

approved