cp's OEIS Frontend

This is a front-end for the Online Encyclopedia of Integer Sequences, made by Christian Perfect. The idea is to provide OEIS entries in non-ancient HTML, and then to think about how they're presented visually. The source code is on GitHub.

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A233862 Prime(k), where k is such that (1 + Sum_{i=1..k} prime(i)^2) / k is an integer.

Original entry on oeis.org

2, 3, 5, 7, 13, 23, 37, 41, 101, 107, 197, 317, 1033, 2029, 2357, 2473, 2879, 5987, 6173, 35437, 56369, 81769, 195691, 199457, 793187, 850027, 1062931, 1840453, 2998421, 4217771, 6200923, 9914351, 10153807, 13563889, 18878099, 60767923, 118825361, 170244929
Offset: 1

Views

Author

Robert Price, Dec 16 2013

Keywords

Comments

a(51) > 1428199016921.
a(67) > 2407033812270611. - Bruce Garner, May 05 2021

Examples

			a(5) = 13, because 13 is the 6th prime and the sum of the first 6 primes^2+1 = 378 when divided by 6 equals 63 which is an integer.
		

Crossrefs

Cf. A085450 (smallest m > 1 such that m divides Sum_{k=1..m} prime(k)^n).

Programs

  • Mathematica
    t = {}; sm = 1; Do[sm = sm + Prime[n]^2; If[Mod[sm, n] == 0, AppendTo[t, Prime[n]]], {n, 100000}]; t (* Derived from A217599 *)
    Module[{nn=9600000},Prime[#]&/@Transpose[Select[Thread[{Range[nn], 1+ Accumulate[ Prime[Range[nn]]^2]}],IntegerQ[Last[#]/First[#]]&]][[1]]] (* Harvey P. Dale, Sep 09 2014 *)
  • PARI
    is(n)=if(!isprime(n),return(0)); my(t=primepi(n),s); forprime(p=2,n,s+=Mod(p,t)^2); s==0 \\ Charles R Greathouse IV, Nov 30 2013

A128169 Numbers k such that k divides 1 + Sum_{j=1..k} prime(j)^5 = 1 + A122103(k).

Original entry on oeis.org

1, 2, 4, 6, 10, 12, 22, 58, 155, 363, 464, 665, 1146, 2870, 3048, 4019, 5931, 8724, 21503, 50439, 67560, 476281, 705570, 4050684, 6956459, 7443590, 10449928, 10799546, 15385564, 17735139, 83325458, 245271750, 255583775, 1395860516, 2921734534, 6255577368, 9050771725, 12062893218, 13689205205, 42254229197, 46440930382
Offset: 1

Views

Author

Alexander Adamchuk, Feb 22 2007, Feb 23 2007

Keywords

Comments

a(52) > 3*10^13. - Bruce Garner, Jun 05 2021
a(53) > 1.2*10^14. - Bruce Garner, Mar 28 2022

Crossrefs

Cf. A085450 (smallest m > 1 such that m divides Sum_{k=1..m} prime(k)^n).

Programs

  • Mathematica
    p = 2; k = 0; s = 1; lst = {}; While[k < 521330000, s = s + p^5; If[Mod[s, ++k] == 0, AppendTo[lst, k]; Print[{k, p}]]; p = NextPrime@ p]

Extensions

a(31) from Sean A. Irvine, Jan 19 2011
a(32)-a(33) from Robert G. Wilson v, Jan 20 2011
a(34)-a(41) from Robert Price, Dec 18 2013

A164576 Integer averages of the set of the first positive squares up to some n^2.

Original entry on oeis.org

1, 11, 20, 46, 63, 105, 130, 188, 221, 295, 336, 426, 475, 581, 638, 760, 825, 963, 1036, 1190, 1271, 1441, 1530, 1716, 1813, 2015, 2120, 2338, 2451, 2685, 2806, 3056, 3185, 3451, 3588, 3870, 4015, 4313, 4466, 4780, 4941, 5271, 5440, 5786, 5963, 6325, 6510
Offset: 1

Views

Author

Keywords

Comments

Integers of the form A000330(k)/k, k listed in A007310. - R. J. Mathar, Aug 20 2009

Examples

			a(1) = 1^2/1 is an integer. The average of the first two squares is (1^2+2^2)/2=5/2, not integer.
The average of the first three squares is (1^2+2^2+3^2)/3=14/3, not integer.
The average of the first five squares is (1^2+2^2+3^2+4^2+5^2)/ 5=11, integer, and constitutes a(2).
		

Crossrefs

Programs

  • Mathematica
    s=0;lst={};Do[a=(s+=n^2)/n;If[Mod[a,1]==0,AppendTo[lst,a]],{n,6!}];lst
    Flatten[Table[{(1 + 3 k) (1 + 4 k), (1 + k) (11 + 12 k)}, {k, 0, 499}]] (* Zak Seidov, Aug 15 2012 *)
    Module[{nn=150,sq},sq=Range[nn]^2;Select[Table[Mean[Take[sq,n]],{n,nn}],IntegerQ]] (* or *) LinearRecurrence[{1,2,-2,-1,1},{1,11,20,46,63},50] (* Harvey P. Dale, Oct 31 2013 *)
  • PARI
    a(n) = 1/4*(12*n^2 - 6*n + (-1)^n*(4*n-1) + 1) \\ Colin Barker, Dec 26 2015

Formula

a(n) = A000330(A007310(n)) / A007310(n) = A175485(A007310(n)). - Jaroslav Krizek, May 28 2010
G.f. ( -x*(1+10*x+7*x^2+6*x^3) ) / ( (1+x)^2*(x-1)^3 ). - R. J. Mathar, Jan 25 2011
a(n) = 1/4*(12*n^2 - 6*n + (-1)^n*(4*n-1) + 1). - Colin Barker, Dec 26 2015

Extensions

Edited by R. J. Mathar, Aug 20 2009

A134125 Integral quotients of partial sums of primes divided by the number of summations.

Original entry on oeis.org

5, 5, 7, 11, 16, 107, 338, 1011, 2249, 22582, 35989, 39167, 61019, 186504, 248776, 367842, 977511, 1790714, 7104697, 15450640, 42428590, 81262621, 232483021, 319278215, 364554172, 419271517, 4432367717, 14591939203, 46911464601, 78572862347, 277369665793, 281386467553
Offset: 1

Views

Author

Enoch Haga, Oct 09 2007

Keywords

Comments

With 1 summation, the partial sum is 2+3 = 5 and 5/1 = 5 is an integer, added to sequence. With 2 summations, the partial sum is 2+3+5 = 10 and 10/2 = 5 is an integer, added to the sequence. After 3 summations, 2+3+5+7 = 17 and 17/3 = 5.6... is not an integer, no contribution to the sequence.
These are all integers of the form A007504(k+1)/k, occurring at k in A134126. Similar to A050248, which looks at A007504(k)/k. - R. J. Mathar, Oct 23 2007

Examples

			a(1) = 5 because 2+3 = 5 and 5/1 = 5, an integral quotient.
a(3) = A007504(5)/4 = 28/4 = 7.
a(4) = A007504(8)/7 = 77/7 = 11.
		

Crossrefs

Programs

  • Mathematica
    With[{nn=50000000},Select[Rest[Accumulate[Prime[Range[nn]]]]/Range[nn-1],IntegerQ]] (* Harvey P. Dale, Jul 25 2013 *)
  • PARI
    lista(pmax) = {my(k = 0, s = 2); forprime(p = 3, pmax, k++; s += p; if(!(s % k), print1(s/k, ", ")));} \\ Amiram Eldar, Apr 30 2024
  • UBASIC
    10 'primes using counters 20 N=3:C=1:R=5:print 2;3,5 30 A=3:S=sqrt(N) 40 B=N\A 50 if B*A=N then N=N+2:goto 30 60 A=A+2:O=A 70 if A<=sqrt(N) then 40 80 C=C+1 90 R=R+N:T=R/C:U=R-N 100 if T=int(T) then print C;U;N;R;T:stop 110 N=N+2:goto 30
    

Formula

a(n) = A007504(k+1)/k where k = A134126(n).

Extensions

a(21) from R. J. Mathar, Oct 23 2007
Edited by R. J. Mathar, Apr 17 2009
a(22)-a(29) from Max Alekseyev, Jan 28 2012
a(30)-a(32) from Amiram Eldar, Apr 30 2024

A125907 Numbers k such that k divides 2^4 + 3^4 + 5^4 + ... + prime(k)^4.

Original entry on oeis.org

1, 2951, 38266951, 3053263643573, 3798632877308897
Offset: 1

Views

Author

Alexander Adamchuk, Feb 04 2007

Keywords

Comments

No more terms to 10^13. - Charles R Greathouse IV, Mar 21 2011
a(4) is less than 10^13 contradicting the previous comment. It was found using the primesieve library by Kim Walisch and gmplib. - Bruce Garner, Feb 26 2021
a(6) > 4*10^15. - Paul W. Dyson, Nov 19 2024

Crossrefs

Cf. A085450 = smallest m > 1 such that m divides Sum_{k=1..m} prime(k)^n.

Programs

  • Mathematica
    a(1) = 1; s = 2^4; Do[s = s + Prime[2n]^4+Prime[2n+1]^4; If[ Mod[s, 2n+1] == 0, Print[2n+1]], {n,1, 20000000}]
  • PARI
    s=0; n=0; forprime(p=2, 4e9, s+=p^4; if(s%n++==0, print1(n", "))) \\ Charles R Greathouse IV, Mar 21 2011

Extensions

a(4) from Bruce Garner, Feb 26 2021
a(5) from Paul W. Dyson, May 09 2024

A128168 Numbers k such that k divides 1 + Sum_{j=1..k} prime(j)^4 = 1 + A122102(k).

Original entry on oeis.org

1, 2, 3, 4, 5, 6, 8, 9, 10, 12, 15, 16, 20, 24, 27, 30, 32, 39, 40, 45, 48, 58, 60, 80, 88, 90, 96, 100, 120, 138, 168, 180, 207, 216, 240, 328, 342, 353, 360, 456, 470, 480, 496, 564, 591, 768, 840, 1040, 1215, 1276, 1355, 1360, 1395, 1440, 1600, 2208, 2576, 2904
Offset: 1

Views

Author

Alexander Adamchuk, Feb 22 2007

Keywords

Comments

a(280) > 5*10^13. - Bruce Garner, Jun 05 2021

Crossrefs

Cf. A085450 (smallest m > 1 such that m divides Sum_{k=1..m} prime(k)^n).

Programs

  • Mathematica
    s = 1; Do[s = s + Prime[n]^4; If[ Mod[s, n] == 0, Print[n]], {n, 17500}]

A164577 Integer averages of the first perfect cubes up to some n^3.

Original entry on oeis.org

1, 12, 25, 45, 112, 162, 225, 396, 507, 637, 960, 1156, 1377, 1900, 2205, 2541, 3312, 3750, 4225, 5292, 5887, 6525, 7936, 8712, 9537, 11340, 12321, 13357, 15600, 16810, 18081, 20812, 22275, 23805, 27072, 28812, 30625, 34476, 36517, 38637, 43120
Offset: 1

Views

Author

Keywords

Comments

Integers of the form A000537(k)/k, created by the k>0 listed in A042965. - R. J. Mathar, Aug 20 2009
Integers of the form (1/4)*n*(n+1)^2 for some n. - Zak Seidov, Aug 17 2009

Examples

			The average of the first cube is 1^3/1=1=a(1).
The average of the first two cubes is (1^3+2^3)/2=9/2, not integer, and does not contribute to the sequence.
The average of the first three cubes is (1^3+2^3+3^3)/3=12, integer, and defines a(2).
		

Crossrefs

Programs

  • Mathematica
    Timing[s=0;lst={};Do[a=(s+=n^3)/n;If[Mod[a,1]==0,AppendTo[lst,a]],{n, 5!}];lst]
    With[{nn=80},Select[#[[1]]/#[[2]]&/@Thread[{Accumulate[Range[ nn]^3],Range[ nn]}],IntegerQ]] (* or *) LinearRecurrence[{1,0,3,-3,0,-3,3,0,1,-1},{1,12,25,45,112,162,225,396,507,637},50] (* Harvey P. Dale, Mar 14 2020 *)

Formula

G.f.: ( x*(1+11*x+13*x^2+17*x^3+34*x^4+11*x^5+6*x^6+3*x^7) ) / ( (1+x+x^2)^3*(x-1)^4 ). - R. J. Mathar, Jan 25 2011

Extensions

Changed comments to examples - R. J. Mathar, Aug 20 2009

A233893 Prime(n), where n is such that (1+sum_{i=1..n} prime(i)^4) / n is an integer.

Original entry on oeis.org

2, 3, 5, 7, 11, 13, 19, 23, 29, 37, 47, 53, 71, 89, 103, 113, 131, 167, 173, 197, 223, 271, 281, 409, 457, 463, 503, 541, 659, 787, 997, 1069, 1279, 1321, 1511, 2203, 2297, 2381, 2423, 3221, 3331, 3413, 3541, 4093, 4327, 5849, 6473, 8291, 9851, 10429, 11177
Offset: 1

Views

Author

Robert Price, Dec 17 2013

Keywords

Comments

a(280) > 1701962315686097. - Bruce Garner, Jun 05 2021

Examples

			a(6) = 13, because 13 is the 6th prime and the sum of the first 6 primes^4+1 = 46326 when divided by 6 equals 7721 which is an integer.
		

Crossrefs

Cf. A085450 (smallest m > 1 such that m divides Sum_{k=1..m} prime(k)^n).

Programs

  • Mathematica
    t = {}; sm = 1; Do[sm = sm + Prime[n]^4; If[Mod[sm, n] == 0, AppendTo[t, Prime[n]]], {n, 100000}]; t (* Derived from A217599 *)
    Module[{nn=1400,t},t=Accumulate[Prime[Range[nn]]^4]+1;Prime[#]&/@ Transpose[Select[Thread[{Range[nn],t}],IntegerQ[#[[2]]/#[[1]]]&]][[1]]](* Harvey P. Dale, Sep 06 2015 *)
  • PARI
    is(n)=if(!isprime(n),return(0)); my(t=primepi(n),s); forprime(p=2,n,s+=Mod(p,t)^4); s==0 \\ Charles R Greathouse IV, Nov 30 2013

A160758 Integer averages of first n nonprime numbers for some n.

Original entry on oeis.org

1, 8, 25, 33, 359, 2948, 3291, 4959, 22350, 33357, 60907, 80962, 8276347, 11856980, 15254419, 176009996, 967538242, 1729774831, 9977169279, 193005936726
Offset: 1

Views

Author

Daniel Tisdale, May 25 2009

Keywords

Comments

A variant of A050248 for nonprimes.
Numbers n such that (1/n)*sum(j=1..n, A018252(j)) is an integer. - Robert G. Wilson v, Jun 05 2009

Examples

			The sum of the first 44 nonprimes is 1452. 1452 / 44 = 33, hence 33 is in the sequence.
		

Crossrefs

Cf. A050248, integer averages of n primes for some n.

Programs

  • Magma
    S:=[]; a:=0; c:=0; for n in [1..40000000] do if not IsPrime(n) then a+:=n; c+:=1; if a mod c eq 0 then Append(~S, a div c); end if; end if; end for; S; // Klaus Brockhaus, Aug 11 2009
  • Mathematica
    lst = {}; s = 0; c = 0; k = 1; While[k < 2700000000, If[ !PrimeQ@k, c++; s = s + k; If[Mod[s, c] == 0, AppendTo[lst, s/c]]]; k++ ]; lst (* Robert G. Wilson v, Jun 05 2009 *)
    a=0;lst={};Do[If[ !PrimeQ[n],m=n;a+=m;If[a/n==IntegerPart[a/n],AppendTo[lst,a/n]]],{n,9!}];lst

Formula

a(n) = A164280(n) / A129749(n).

Extensions

a(6) - a(16) from Robert G. Wilson v, Jun 05 2009
a(17) - a(19) from Donovan Johnson, Sep 16 2009
Edited by N. J. A. Sloane, May 11 2010
a(20) from Donovan Johnson, May 20 2010

A223937 a(n) is the sum of the cubes of the first A122140(n) primes.

Original entry on oeis.org

8, 4696450, 7024453131396, 17761740387522, 155912686127038650, 87598780898450312031408, 2147216863131055036604400, 2908950240914054780101441371333254159676520, 384422969812280951687876430655304031054262132, 6187047308209705064673104196645071104957480508
Offset: 1

Views

Author

Robert Price, Mar 29 2013

Keywords

Crossrefs

Cf. A085450 (smallest m > 1 that divides Sum_{k=1..m} prime(k)^n), A122140.

Extensions

Title corrected by Hugo Pfoertner, Feb 09 2021
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