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A096278
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Sums of successive sums of successive sums of successive primes.
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5
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33, 50, 72, 96, 120, 144, 172, 206, 240, 274, 308, 336, 364, 402, 444, 480, 514, 548, 578, 610, 648, 692, 742, 786, 816, 840, 864, 900, 960, 1024, 1070, 1108, 1152, 1196, 1236, 1278, 1320, 1362, 1404, 1444, 1488, 1530, 1560, 1592, 1650, 1728, 1790, 1824
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OFFSET
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1,1
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COMMENTS
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If we consider the m-fold iterated "take sums of successive terms" operation acting on the primes, then for all m >= 1, the first term is always odd (and the only odd term); it is prime for m=1, 2, 4, 8, 21, 24, 27, 31, 40, 98,..., but not for m=3 (the present sequence). [Edited by M. F. Hasler, Jun 02 2017]
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LINKS
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FORMULA
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Let f(n) = prime(n) + prime(n+1) f1(n) = f(n)+f(n+1) : SS of order 1 Then f2(n) = f1(n)+f1(n) : SS of order 2 is the general term of this sequence.
a(n) = prime(n)+3*prime(n+1)+3*prime(n+2)+prime(n+3). - Robert Israel, Dec 28 2022
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EXAMPLE
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The first two terms of SS order 1 is 13 and 20. 13+20 = 33 the first term of the sequence.
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MAPLE
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Ss:= L -> L[1..-2]+L[2..-1]:
(Ss@@3)([seq(ithprime(i), i=1..100)]); # Robert Israel, Dec 28 2022
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MATHEMATICA
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Nest[ListConvolve[{1, 1}, #]&, Prime[Range[100]], 3] (* Paolo Xausa, Oct 31 2023 *)
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PROG
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(PARI) g(n) = for(x=1, n, print1(f2(x)", ")) f(n) = return(prime(n)+prime(n+1)) f1(n) = return(f(n)+f(n+1)) f2(n) = return(f1(n)+f1(n+1))
(PARI) A096278(n, m=3)=for(k=0, m, prime(n+k)*binomial(m, k)) \\ or, to get a list:
A096278_vec(Nmax, m=3, v=primes(Nmax+m))=sum(k=0, m, binomial(m, k)*v[1+k, k-1-m]) \\ Alternatively, do m times v=v[^1]+v[^-1]. - M. F. Hasler, Jun 02 2017
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CROSSREFS
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KEYWORD
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easy,nonn
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AUTHOR
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STATUS
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approved
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