A172524 Partial sums of Iccanobif numbers A001129.

0, 1, 2, 4, 7, 12, 20, 33, 72, 196, 710, 1546, 2599, 6738, 19553, 80688, 185625, 978142, 2432840, 12112678, 29466988, 39202128, 40962878, 41948928, 42570288, 42684103, 43265540, 44518036, 52194742, 65214030, 159581828, 337649208

Offset: 0

Author

Jonathan Vos Post, Feb 06 2010

Keywords

• base
• easy
• nonn

Comments

The only primes in this sequence are: 2, 7 and 19553. The squares in this sequence begin: 0, 1, 4, 196.

Examples

			a(14) = 0 + 1 + 1 + 2 + 3 + 5 + 8 + 13 + 39 + 124 + 514 + 836 + 1053 + 4139 + 12815 = 19553 is prime. a(31) = 0 + 1 + 1 + 2 + 3 + 5 + 8 + 13 + 39 + 124 + 514 + 836 + 1053 + 4139 + 12815 + 61135 + 104937 + 792517 + 1454698 + 9679838 + 17354310 + 9735140 + 1760750 + 986050 + 621360 + 113815 + 581437 + 1252496 + 7676706 + 13019288 + 94367798 + 178067380.
		

Crossrefs

Cf. A000040, A000045, A001129, A004086, A014258-A014260.

Programs

Mathematica

nxt[{a_,b_}]:={b,Total[FromDigits/@Reverse/@IntegerDigits[{a,b}]]};Accumulate[ Transpose[NestList[nxt,{0,1},40]][[1]]] (* Harvey P. Dale, Apr 04 2015 *)

Formula

a(n) = SUM[i=0..n] A001129(i) = SUM[i=0..n] {a(0) = 0, a(1) = 1, a(i+2) = R(a(i)) + R(a(i+1))} = SUM[i=0..n] A001129(i) = SUM[i=1..n] {a(0) = 0, a(1) = 1, a(i+2) = A004086(a(i)) + A004086(a(i+1))}.