A082548 a(n) is the number of values of k such that k can be expressed as the sum of distinct primes with largest prime in the sum equal to prime(n).

1, 2, 4, 7, 12, 23, 36, 53, 72, 95, 124, 155, 192, 233, 276, 323, 376, 435, 496, 563, 634, 707, 786, 869, 958, 1055, 1156, 1259, 1366, 1475, 1588, 1715, 1846, 1983, 2122, 2271, 2422, 2579, 2742, 2909, 3082, 3261, 3442, 3633, 3826, 4023, 4222, 4433, 4656, 4883

Offset: 1

Author

Naohiro Nomoto, May 02 2003

Keywords

• easy
• nonn

Comments

Surprisingly, except for the initial term, the first differences of this sequence is the sequence of primes with 7 omitted. - John W. Layman, Feb 25 2012

Also number of k that can be expressed as a signed sum of the first n-1 primes. - Seiichi Manyama, Oct 01 2019

Examples

			For n=4; 7 is the 4th prime. 7 = 7, 9 = 2+7, 10 = 3+7, 12 = 5+7 = 2+3+7, 14 = 2+5+7, 15 = 3+5+7, 17 = 2+3+5+7. Values of m are 7 and 9,10,12,14,15,17. so a(4)=7.
From _Seiichi Manyama_, Oct 01 2019: (Start)
7       =  7, so 7*2         = 14 = 24-10 = 24+(-2-3-5).
2+7     =  9, so (2+7)*2     = 18 = 24- 6 = 24+( 2-3-5).
3+7     = 10, so (3+7)*2     = 20 = 24- 4 = 24+(-2+3-5).
5+7     = 12, so (5+7)*2     = 24 = 24+ 0 = 24+(-2-3+5).
2+5+7   = 14, so (2+5+7)*2   = 28 = 24+ 4 = 24+( 2-3+5).
3+5+7   = 15, so (3+5+7)*2   = 30 = 24+ 6 = 24+(-2+3+5).
2+3+5+7 = 17. so (2+3+5+7)*2 = 34 = 24+10 = 24+( 2+3+5). (End)
From _Seiichi Manyama_, Oct 02 2019: (Start)
Let b(n) be the number of k (>=0) that can be expressed as the sum of distinct primes with largest prime in the sum not greater than prime(n).
n |b(n)|            |
--+----+------------+--------------------------------------
4 | 12 |  0         | 11
  |    |  2         | 13 =  2+11
  |    |  3         | 14 =  3+11
  |    |  5         | 16 =  5+11
  |    |  7         | 18 =  7+11
  |    |  8 = 3+5   | 19 =  8+11 = (3+5)+11
  |    |  9 = 17-8  | 20 =  9+11 = (2+3+5+7)-(3+5)+11
  |    | 10 = 17-7  | 21 = 10+11 = (2+3+5+7)-7    +11
  |    | 12 = 17-5  | 23 = 12+11 = (2+3+5+7)-5    +11
  |    | 14 = 17-3  | 25 = 14+11 = (2+3+5+7)-3    +11
  |    | 15 = 17-2  | 26 = 15+11 = (2+3+5+7)-2    +11
  |    | 17 = 17-0  | 28 = 17+11 = (2+3+5+7)      +11
--+----+------------+--------------------------------------
5 | 23 |  0         | 13
  |    |  2         | 15 =  2+13
  |    |  3         | 16 =  3+13
  |    |  5         | 18 =  5+13
  |    |  7         | 20 =  7+13
  |    |  8 = 3+5   | 21 =  8+13 = (3+5)  +13
  |    |  9 = 2+7   | 22 =  9+13 = (2+7)  +13
  |    | 10 = 2+3+5 | 23 = 10+13 = (2+3+5)+13
  |    | 11         | 24 = 11+13
  |    | ...        | ...
  |    | 17 = 28-11 | 30 = 17+13 = (2+3+5+7+11)-11     +13
  |    | 18 = 28-10 | 31 = 18+13 = (2+3+5+7+11)-(2+3+5)+13
  |    | 19 = 28- 9 | 32 = 19+13 = (2+3+5+7+11)-(2+7)  +13
  |    | 20 = 28- 8 | 33 = 20+13 = (2+3+5+7+11)-(3+5)  +13
  |    | 21 = 28- 7 | 34 = 21+13 = (2+3+5+7+11)- 7     +13
  |    | 23 = 28- 5 | 36 = 23+13 = (2+3+5+7+11)- 5     +13
  |    | 25 = 28- 3 | 38 = 25+13 = (2+3+5+7+11)- 3     +13
  |    | 26 = 28- 2 | 39 = 26+13 = (2+3+5+7+11)- 2     +13
  |    | 28 = 28- 0 | 41 = 28+13 = (2+3+5+7+11)        +13
--+----+------------+-------------------------------------
...
b(n) = Sum_{k=1..n} prime(k) + 1 - 3*2 = A007504(n) - 5 for n>3.
So a(n) = b(n-1) = A007504(n-1) - 5 for n>4. (End)
		

Links

• Seiichi Manyama, Table of n, a(n) for n = 1..10000

Crossrefs

Cf. A007504, A082533, A082534, A327467.

Programs

PARI

limit = 70; M = sum(i = 1, limit, prime(i)); v = vector(M); primeSum = 0; forprime (n = 1, prime(limit), count = 1; forstep (i = primeSum, 1, -1, if (v[i], count = count + 1; v[i + n] = 1)); v[n] = 1; print(count); primeSum = primeSum + n)

Formula

a(n) = A007504(n-1) - 5 for n > 4. - Seiichi Manyama, Oct 02 2019

Extensions

More terms from David Wasserman, Sep 16 2004