A055233 -id:A055233 - cp's OEIS frontend

A074036 Sum of the primes from the smallest prime factor of n to the largest prime factor of n.

0, 2, 3, 2, 5, 5, 7, 2, 3, 10, 11, 5, 13, 17, 8, 2, 17, 5, 19, 10, 15, 28, 23, 5, 5, 41, 3, 17, 29, 10, 31, 2, 26, 58, 12, 5, 37, 77, 39, 10, 41, 17, 43, 28, 8, 100, 47, 5, 7, 10, 56, 41, 53, 5, 23, 17, 75, 129, 59, 10, 61, 160, 15, 2, 36, 28, 67, 58, 98, 17, 71, 5, 73, 197, 8

Offset: 1

Jason Earls, Sep 15 2002

Obviously if n is prime then a(n) = n. However, there are composite values of n such that a(n) = n, such as 10 and 155. - Alonso del Arte, May 30 2017

			a(14) = 17 because 14 = 2 * 7 and 2 + 3 + 5 + 7 = 17.

Alois P. Heinz, Table of n, a(n) for n = 1..20000
Erich Friedman, What's Special About This Number? Entry for 155.

Cf. A055233, A169802, A169804.

f:=proc(n) local i,t1,t2,t3,t4,t5,t6; if n<=1 then RETURN(0) else
t1:=ifactors(n); t2:=t1[2]; t3:=nops(t2); t4:=0; t5:=pi(t2[1][1]); t6:=pi(t2[t3][1]);
for i from t5 to t6 do t4:=t4+ithprime(i); od; RETURN(t4); fi; end; # N. J. A. Sloane, May 24 2010
# second Maple program:
s:= proc(n) option remember; `if`(n<1, 0, ithprime(n)+s(n-1)) end:
a:= proc(n) option remember; uses numtheory; `if`(n<2, 0, (m->
      s(pi(max(m)))-s(pi(min(m))-1))(factorset(n)))
    end:
seq(a(n), n=1..100);  # Alois P. Heinz, Nov 24 2021

sp[n_]:=With[{fi=FactorInteger[n][[All,1]]},Total[Prime[Range[ PrimePi[ fi[[1]]],PrimePi[fi[[-1]]]]]]]; Join[{0},Array[sp,80,2]] (* Harvey P. Dale, Dec 22 2017 *)

a(n) = if (n==1, 0, my(f = factor(n), s = 0); forprime(p=f[1,1], f[#f~,1], s += p); s); \\ Michel Marcus, May 31 2017

Given p prime and k > 0, a(p^k) = p. - Alonso del Arte, May 30 2017

A055514 Composite numbers that are the sum of consecutive prime numbers and are divisible by the first and last of these primes.

Original entry on oeis.org

10, 39, 155, 371, 10225245560, 2935561623745, 454539357304421, 7228559051256366318, 1390718713078158117206

Offset: 1

Jud McCranie, Jul 03 2000

Composite n such that n = p_1 + p_2 + ... + p_k where the p_i are consecutive primes and n is divisible by p_1 and p_k.
Problem proposed by Carlos Rivera, who found the first 4 terms.
No more terms below 10^22. - Michael Beight, Jul 22 2012
In subsequence A055233 the first and last term of the sum must also be its smallest and largest prime factor. Therefore a(5) (cf. first EXAMPLE) is not in that sequence, since it has smaller factors 2^3*5. - M. F. Hasler, Nov 21 2021

			503 + 509 + 521 + ... + 508213 = 10225245560, which is divisible by 503 and 508213. - _Manuel Valdivia_, Nov 17 2011
From _Michael Beight_, Jul 22 2012: (Start)
a(8) = 7228559051256366318 = 73 + ... + 18281691653;
a(9) = 1390718713078158117206 = 370794889 + ... + 267902967061. (End)

C. Rivera, Puzzle

Subsequence of A050936.

Cf. A055233.

Module[{nn=200},Table[Total/@Select[Partition[Prime[Range[10000]],n,1],scpQ],{n,2,nn}]]//Flatten (* The program generates the first four terms of the sequence. *)
(* Harvey P. Dale, Oct 22 2022 *)

S=vector(N=50000); s=0; i=1; forprime(p=2,oo, S[i++]=s+=p; for(j=1,i-2, (s-S[j])%p || (s-S[j])%prime(j)|| print1(s-S[j]",")|| break)) \\ gives a(1..5), but too slow to go beyond. - M. F. Hasler, Nov 21 2021

a(7) from Donovan Johnson, Jun 19 2008

a(8) and a(9) from Michael Beight, Jul 22 2012

A086447 a(n) = the least k such that prime(n+1)+prime(n+2)+...+prime(n+k) is a multiple of prime(n).

Original entry on oeis.org

2, 2, 6, 6, 6, 6, 4, 8, 4, 30, 7, 31, 37, 67, 22, 60, 46, 38, 69, 13, 65, 76, 19, 163, 9, 52, 100, 84, 66, 136, 66, 119, 33, 79, 47, 76, 187, 214, 37, 96, 461, 111, 62, 189, 510, 37, 256, 121, 130, 132, 144, 481, 64, 195, 53, 47, 136, 90, 194, 318, 526, 151, 788, 1542

Offset: 1

Zak Seidov, Jul 20 2003

Conjecture: a(n) exists for every n.

			a(3)=6 because prime(3)=5 divides 7+11+13+17+19+23 = 90.

Charles R Greathouse IV, Table of n, a(n) for n = 1..10000

Cf. A055233, A055514, A086448.

bb={}; Do[s0=Prime[n0]; s=0; Do[s+=Prime[n]; If[IntegerQ[s/s0], bb=Append[bb, n-n0]; Break[]], {n, n0+1, 8000}], {n0, 1, 100}]; bb
sncp[n_]:=Module[{p=Prime[n],k=n+1,t},t=Prime[k];While[!Divisible[ t, p], k++;t=t+Prime[k]];k-n]; Array[sncp,100]  (* Harvey P. Dale, May 21 2017 *)

a(n)=my(p = prime(n), sp = nextprime(p+1), lp = sp, nb = 1); while (sp % p, lp = nextprime(lp+1); nb++; sp += lp); nb; \\ Michel Marcus, May 21 2017

a(n, p=prime(n))=my(s, k); forprime(q=p+1, , s+=q; k++; if(s%p==0, return(k))) \\ Charles R Greathouse IV, May 21 2017

Edited by Don Reble, Nov 10 2005

A086448 a(n) = the least integer of the form [prime(n+1)+prime(n+2)+...+prime(n+k)]/prime(n).

Original entry on oeis.org

4, 4, 18, 16, 12, 12, 6, 16, 6, 102, 11, 93, 119, 345, 48, 240, 138, 100, 263, 19, 227, 282, 31, 1071, 11, 126, 386, 278, 184, 642, 164, 445, 55, 213, 89, 190, 895, 1120, 61, 258, 4629, 323, 122, 789, 5226, 59, 1292, 325, 364, 374, 430, 3939, 118, 695, 87, 73, 358

Offset: 1

Zak Seidov, Jul 20 2003

It seems that a(n) exists for any n.
Among first 1000 terms, the largest term is a(793) = 1807606, with p = prime(793) = 6079, and 6079*1807606 = the sum of 42840 consecutive primes after p. - Zak Seidov, Nov 07 2014
Among first 10000 terms, the largest term is a(9349) = 30376745, with p = prime(9349) = 97159, and p*(9349) = the sum of 629543 consecutive primes after p: 2951374167455 = sum(prime(k), k = 9349 + 1..9349 + 629543) - Zak Seidov, Feb 21 2015

			a(3)=18 because prime(3)=5 and (7+11+13+17+19+23)/5 = 18.

Zak Seidov, Table of n, a(n) for n = 1..10000

Cf. A055233, A055514, A086447.

bb={}; Do[s0=Prime[n0]; s=0; Do[If[IntegerQ[ss0=(s+=Prime[n])/s0], bb=Append[bb, ss0]; Break[]], {n, n0+1, 8000}], {n0, 1, 10}]; bb

Edited by Don Reble, Nov 10 2005

A169802 Numbers k equal to A074036(k) = sum of primes from least to largest prime factor.

Original entry on oeis.org

2, 3, 5, 7, 10, 11, 13, 17, 19, 23, 29, 31, 37, 39, 41, 43, 47, 53, 59, 61, 67, 71, 73, 79, 83, 89, 97, 101, 103, 107, 109, 113, 127, 131, 137, 139, 149, 151, 155, 157, 163, 167, 173, 179, 181, 191, 193, 197, 199, 211, 223, 227, 229, 233, 239, 241, 251, 257, 263, 269, 271

Offset: 1

N. J. A. Sloane, May 24 2010

nonn

Union of primes (A000040) and A055233.

Erich Friedman, What's Special About This Number?

Cf. A055233 (this without primes), A074036 (sum of primes from spf(n) to gpf(n)).

Cf. A020639 (spf: smallest prime factor), A006530 (gpf: greatest prime factor).

upto=500;a={};Do[f=Map[First,FactorInteger[k]];If[k==Total[Select[Range[First[f],Last[f]],PrimeQ]],AppendTo[a,k]],{k,upto}]; a (* Paolo Xausa, Nov 27 2021 *)

select( {is_A169802(n,f=factor(n)[,1])=n>1&&n==vecsum(primes([f[1],f[#f]]))}, [1..222]) \\ M. F. Hasler, Nov 24 2021

A303556 Numbers equal to the sum of the numbers between two of their consecutive divisors.

Original entry on oeis.org

490, 55930, 98648, 222560, 396550, 584988, 838448, 1173102, 2345720, 2855660, 4150120, 4781502, 5557300, 6072460, 6115122, 6688416, 6715280, 9390290, 9486950, 11691498, 12704510, 13331240, 16035760, 17325700, 19377050, 20055070, 20859410, 29651748, 34516160, 35040352

Offset: 1

Paolo P. Lava, Apr 26 2018

nonn

If also the two consecutive divisors were added to the sum, the first terms would be 18, 55120, 1034540, 1386350, 1675960, ...

			a(1) = 490 because 14 and 35 are two consecutive divisors of 490 and the sum of the numbers from 15 to 34 is equal to 490 itself.
a(7) = 838448 because 1807 and 2224 are two consecutive divisors of 838448 and the sum of the numbers from 1808 to 2223 is equal to 838448 itself.

Cf. A055233.

with(numtheory): P:=proc(q) local a,k,n;
for n from 1 to q do if not isprime(n) then a:=sort([op(divisors(n))]);
for k from 1 to tau(n)-1 do if n=((a[k+1]-1)*a[k+1]-a[k]*(a[k]+1))/2
then print(n); break; fi; od; fi; od; end: P(10^9);

Select[Range[351*10^5],MemberQ[Total[Range[#[[1]]+1,#[[2]]-1]]&/@Partition[ Divisors[ #],2,1],#]&] (* Harvey P. Dale, Feb 14 2023 *)

isok(n) = my(d=divisors(n)); vecsearch(vecsort(vector(#d-1, k, ((d[k+1]-1)*d[k+1]-d[k]*(d[k]+1))/2),,8), n); \\ Michel Marcus, Apr 27 2018

a(10)-a(30) from Giovanni Resta, Apr 27 2018

A323441 Numbers equal to the sum of the numbers between two of their divisors.

Original entry on oeis.org

12, 30, 60, 90, 108, 180, 234, 240, 390, 408, 420, 462, 480, 490, 630, 756, 840, 880, 900, 945, 1122, 1218, 1248, 1430, 1500, 1512, 1560, 1584, 1998, 2070, 2100, 2310, 2460, 2520, 2652, 2660, 2970, 3306, 3330, 3528, 3780, 3960, 4004, 4032, 4134, 4140, 4275, 4788

Offset: 1

Paolo P. Lava, Jan 15 2019

A303556 is a subsequence of this sequence.

			Divisors of 12 are 1, 2, 3, 4, 6, 12 and the sum of the numbers between 2 and 6 is 3 + 4 + 5 = 12.
Divisors of 108 are 1, 2, 3, 4, 6, 9, 12, 18, 27, 36, 54, 108 and the sum of the numbers between 9 and 18 is 10 + 11 + 12 + 13 + 14 + 15 + 16 + 17 = 108.

Paolo P. Lava, Table of n, a(n) for n = 1..1000

Cf. A055233, A303556.

with(numtheory): P:=proc(n) local a, j, k; a:=sort([op(divisors(n))]);
for j from 1 to nops(a)-1 do for k from j+1 to nops(a) do
if n=((a[k]-1)*a[k]-a[j]*(a[j]+1))/2 then RETURN(n); fi;
od; od; end: seq(P(i), i=2..10^4);

Select[Range[5000], MemberQ[Union@ Map[Total@ Range[#1 + 1, #2 - 1] & @@ # &, Subsets[Divisors@ #, {2}]], #] &] (* Michael De Vlieger, Jan 18 2019 *)

A074195 Sum of the primes from smallest prime factor of n to the largest prime factor of n = largest difference between consecutive divisors of n (ordered by size).

Original entry on oeis.org

4, 20, 1278, 1339, 11074308238, 19096347067

Offset: 1

Jason Earls, Sep 19 2002

10^11 < a(7) <= 8912510129422438. - Giovanni Resta, May 13 2016

Cf. A060681, A074036, A055233, A055514.

Select[Range[2, 2000], (p = First /@ FactorInteger[#]; #-#/p[[1]] == Sum[ Prime[i], {i, PrimePi@ p[[1]], PrimePi@ p[[-1]]}]) &] (* Giovanni Resta, May 13 2016 *)

isok(n) = {pf = factor(n)[,1]; my(pmin = vecmin(pf)); s = 0; forprime(p = pmin, vecmax(pf), s += p); s == n - n/pmin;} \\ Michel Marcus, Feb 03 2014

a(5)-a(6) from Giovanni Resta, May 13 2016

A200721 Product of two nonconsecutive primes p and q that divides the sum of primes between p and q, exclusively.

Original entry on oeis.org

26, 1133, 20309, 51159, 3246905, 28673661, 5201685791

Offset: 1

Manuel Valdivia, Nov 21 2011

nonn

Prime p is approximately q/((2*log(q)-1)*k), for k = 1, 1, 3, 307, 5041, 102378,..(quotients).

a(8) > 2*10^10. 3235398421447 is also a term. - Donovan Johnson, Nov 23 2011

			51159 = 3*17053, (5+ ... +17047)/51159 = 307.

Cf. A055514, A055233, A086447, A086448, A001358.

ss[n_] := Module[{f = Transpose[FactorInteger[n]], p, q, s}, If[f[[2]] == {1, 1}, {p, q} = PrimePi[f[[1]]]; s = Total[Table[Prime[i], {i, p + 1, q - 1}]]; s != 0 && Mod[s, n] == 0, False]]; Select[Range[2, 21000], ss] (* T. D. Noe, Nov 21 2011 *)

a(7) from Donovan Johnson, Nov 23 2011

A309238 Non-prime-square numbers equal to the sum of squares of consecutive primes from the least prime factor to the largest prime factor.

Original entry on oeis.org

315797, 9634877

Offset: 1

Miroslav Kures, Jul 17 2019

So-called 2-straddled numbers; 1-straddled numbers are in A055233.

a(3) > 7*10^14, if it exists. - Giovanni Resta, Jul 18 2019

			9634877 = 7 * 41 * 59 * 569 = 7^2 + 11^2 + 13^2 + ... + 569^2.

M. Kures, Straddled numbers: numbers equal to the sum of powers of consecutive primes from the least prime factor to the largest prime factor, Notes on Number Theory and Discrete Mathematics, 2019, vol. 25, no. 2, pp. 8-15. ISSN: 1310-5132.

Cf. A055233.

isok(n) = if (isprimepower(n) != 2, if (n>1, my(f=factor(n)[,1], s=0); forprime(p=vecmin(f), vecmax(f), s+= p^2); s == n)); \\ Michel Marcus, Jul 18 2019

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A074036 Sum of the primes from the smallest prime factor of n to the largest prime factor of n.

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A055514 Composite numbers that are the sum of consecutive prime numbers and are divisible by the first and last of these primes.

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A086447 a(n) = the least k such that prime(n+1)+prime(n+2)+...+prime(n+k) is a multiple of prime(n).

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A086448 a(n) = the least integer of the form [prime(n+1)+prime(n+2)+...+prime(n+k)]/prime(n).

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A169802 Numbers k equal to A074036(k) = sum of primes from least to largest prime factor.

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A303556 Numbers equal to the sum of the numbers between two of their consecutive divisors.

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A323441 Numbers equal to the sum of the numbers between two of their divisors.

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