A133549
Sum of the fourth powers of the first n odd primes.
Original entry on oeis.org
81, 706, 3107, 17748, 46309, 129830, 260151, 539992, 1247273, 2170794, 4044955, 6870716, 10289517, 15169198, 23059679, 35177040, 49022881, 69174002, 94585683, 122983924, 161934005, 209392326, 272134567, 360663848, 464724249, 577275130
Offset: 1
a(2)=706 because 3^4 + 5^4 = 706.
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a:=proc (n) options operator, arrow: add(ithprime(j)^4, j=2..n+1) end proc: seq(a(n),n=1..26); # Emeric Deutsch, Oct 02 2007
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c = 4; a = {}; b = 0; Do[b = b + Prime[n]^c; AppendTo[a, b], {n, 2, 1000}]; a
A263170
a(n) = (Sum_{k=1..n} prime(k))^3 - (Sum_{k=1..n} prime(k)^3).
Original entry on oeis.org
0, 90, 840, 4410, 20118, 64890, 186168, 440730, 972030, 2094330, 4013850, 7512570, 13279548, 21906810, 34902498, 54772410, 84444690, 124785210, 181983378, 259292154, 358930146, 492406650, 664548816, 889272570, 1186319550, 1559209530, 2012668266, 2568943290, 3232452450, 4031692410
Offset: 1
For n = 2, a(2) = (2 + 3)^3 - (2^3 + 3^3) = 90.
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A263170 := proc(n)
su := add(ithprime(i),i=1..n) ;
su3 := add(ithprime(i)^3,i=1..n) ;
su^3-su3 ;
end proc: # R. J. Mathar, Oct 21 2015
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Table[Sum[Prime@ k, {k, n}]^3 - Sum[Prime[k]^3, {k, n}], {n, 30}] (* Michael De Vlieger, Oct 19 2015 *)
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a(n) = sum(k=1, n, prime(k))^3 - sum(k=1, n, prime(k)^3);
A097881
Decimal expansion of the sum from 1 to infinity of fraction sequence with numerator triangular numbers and denominator sum of prime cubes.
Original entry on oeis.org
2, 9, 4, 3, 9, 4, 2, 7
Offset: 0
A118219
Smallest number k>1 such that Sum_{i=1..k} Prime[i]^n divides Product_{i=1..k} Prime[i]^n.
Original entry on oeis.org
3, 30, 17, 248, 515, 49682
Offset: 1
a(1) = 3 because 2 + 3 + 5 = 10 divides 2*3*5 = 30 but 2 + 3 = 5 does not divide 2*3 = 6.
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f[n_] := Block[{k = 2, p = 2, s = 2^n}, While[p = p*Prime@ k; s = s + Prime@ k^n; PowerMod[p, n, s] != 0, k++ ]; k]; Do[ Print@ f@n, {n, 10}] (* Robert G. Wilson v *)
A122124
Numbers n such that 25 divides Sum[ Prime[k]^n, {k,1,n}].
Original entry on oeis.org
3, 5, 7, 11, 15, 19, 23, 25, 27, 31, 35, 39, 43, 45, 47, 51, 55, 59, 63, 65, 67, 71, 75, 79, 83, 85, 87, 91, 95, 99, 103, 105, 107, 111, 115, 119, 123, 125, 127, 131, 135, 139, 143, 145, 147, 151, 155, 159, 163, 165, 167, 171, 175, 179, 183, 185, 187, 191, 195, 199
Offset: 1
There are 25 primes p < 100, p(n) = {2, 3, 5, 7, 11, 13, 17, 19, 23, 29, 31, 37, 41, 43, 47, 53, 59, 61, 67, 71, 73, 79, 83, 89, 97}.
a(1) = because 25 divides Sum[p(n)^3,{n,1,25}] = 2^3 + 3^3 + ... + 89^3 + 97^3 = A098999[25] and does not divide Sum[p(n)^1,{n,1,25}] = A007504[25] and Sum[p(n)^2,{n,1,25}] = A024450[25].
The next a(2) = 5 because 25 divides Sum[p(n)^5,{n,1,25}] = A122103[25] and does not divide Sum[p(n)^4,{n,1,25}] = A122102[25].
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Select[Range[300],IntegerQ[Sum[ Prime[k]^#1, {k,1,25}]/25]&]
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for(n=1,100,if(sum(k=1,25,prime(k)^n)%25==0,print1(n,",")));
print;print("Alternative method not using primes:");
for(n=1,100,m=(n-1)%6;print1((n-m)*3+(n-m+if(m>1,(m-1)*12-1,m*6-1))/3,",")) \\ K. Spage, Oct 23 2009
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