A117061 Numbers n such that a(n) = (s(n-1))^2+n, with a(1) = 1.

1, 3, 12, 13, 21, 15, 43, 57, 153, 91, 111, 21, 22, 30, 24, 52, 66, 162, 100, 21, 30, 31, 39, 168, 250, 75, 171, 109, 129, 174, 175, 201, 42, 70, 84, 180, 118, 138, 183, 184, 210, 51, 79, 300, 54, 127, 147, 192, 193, 219, 195, 277, 309, 198, 379, 417, 201, 67, 228, 204

Offset

1,2

Comments

From a puzzle : 1 3 12 13 21 15 43 .?. 153 answer : 57 Each number can be found by squaring the sum of the digits of the previous number and then adding the indexnumber.

Links

Harvey P. Dale, Table of n, a(n) for n = 1..1000

Formula

a(n) = s(n-1))^2+n, where s(n) stands for the sum of the digits of a(n).

Mathematica

Join[{k = 1}, Table[k = (Total[IntegerDigits[k]])^2 + n, {n, 2, 60}]] (* Jayanta Basu, Jul 13 2013 *)
nxt[{n_, a_}]:={n+1, Total[IntegerDigits[a]]^2+n+1}; Transpose[NestList[nxt, {1, 1}, 60]][[2]] (* Harvey P. Dale, Dec 25 2013 *)

Crossrefs

Sequence in context: A073542 A063444 A353439 * A341799 A089919 A222711

Adjacent sequences: A117058 A117059 A117060 * A117062 A117063 A117064

Keyword

base,easy,nonn

Author

Luc Stevens (lms022(AT)yahoo.com), Apr 16 2006

Status

approved