Ed Smiley - cp's OEIS frontend

A226502 Let P(k) denote the k-th prime (P(1)=2, P(2)=3 ...); a(n) = P(n+1)P(n+3) - P(n)P(n+2).

11, 34, 36, 96, 60, 144, 160, 162, 360, 198, 320, 336, 352, 494, 460, 720, 378, 560, 718, 450, 972, 1020, 938, 1002, 816, 420, 864, 1752, 960, 2596, 810, 2204, 576, 2404, 1220, 1606, 1980, 1694, 1420, 2876, 744, 2694, 780, 3160, 2810, 3520, 3170, 1824, 1840, 1422, 3836

Offset: 1

Ed Smiley, Jun 09 2013

Differences of the products of alternate primes.

Harvey P. Dale, Table of n, a(n) for n = 1..1000

First differences of A090076.

#[[2]]#[[4]]-#[[1]]#[[3]]&/@Partition[Prime[Range[60]],4,1] (* Harvey P. Dale, Jul 14 2025 *)

p=2;q=3;r=5;forprime(s=7,1e2,print1(q*s-p*r", ");p=q;q=r;r=s) \\ Charles R Greathouse IV, Jun 10 2013

a(n) >> n log n and this is probably sharp: on Dickson's conjecture there are infinitely many a(n) < kn log n for any k > 4. The constant 4 comes from 8 + 2 - 6 - 0 n the prime quadruplet (p+0, p+2, p+6, p+8). On Cramér's conjecture a(n) = O(n log^3 n). Unconditionally a(n) << n^1.525 log n. - Charles R Greathouse IV, Jun 10 2013

A182664 a(n) = A088828(n) + A157502(n).

Original entry on oeis.org

5, 11, 15, 21, 25, 29, 35, 39, 43, 47, 53, 57, 61, 65, 69, 75, 79, 83, 87, 91, 95, 101, 105, 109, 113, 117, 121, 125, 131, 135, 139, 143, 147, 151, 155, 159, 165, 169, 173, 177, 181, 185, 189, 193, 197, 203, 207, 211, 215, 219, 223, 227, 231, 235, 239, 245

Offset: 1

Ed Smiley, Dec 23 2012

nonn

Cf. A088828 and A157502. Also alternate generation formula related to A000217, A010054.

Module[{nn=60,tb},tb=2*Accumulate[Table[If[IntegerQ[(Sqrt[8n+1]-1)/2],1,0],{n,nn}]];Join[{5},Table[1+4i+tb[[i-1]],{i,2,nn}]]] (* Harvey P. Dale, Jul 22 2013 *)

Let T(n) be 1 if positive n is a triangular number, else 0; we also define T(0) as 0.  Then we can also write this sequence as 1 + 4*n + 2*[T(1) + T(2) ... T(n-1)]. (Other than the special definition for T(0), T(n) is essentially A010054.)
The sequence can also be seen visually as
+ 4
+ 4 + 6
+ 4 + 6 + 4
+ 4 + 6 + 4 + 4
+ 4 + 6 + 4 + 4 + 6
+ 4 + 6 + 4 + 4 + 6 + 4
+ 4 + 6 + 4 + 4 + 6 + 4 + 4
+ 4 + 6 + 4 + 4 + 6 + 4 + 4 + 4
+ 4 + 6 + 4 + 4 + 6 + 4 + 4 + 4 + 6...

A001099 a(n) = n^n - a(n-1), with a(1) = 1.

Original entry on oeis.org

1, 3, 24, 232, 2893, 43763, 779780, 15997436, 371423053, 9628576947, 275683093664, 8640417354592, 294234689237661, 10817772136320355, 427076118244539020, 18019667955465012596, 809220593930871751581, 38537187481365665823843, 1939882468178947923300136

Offset: 1

Ed Smiley

nonn

T. D. Noe, Table of n, a(n) for n = 1..100

Cf. A001923.

Abs[Table[Sum[k^k*(-1)^(k+1),{k,1,n}],{n,1,30}]] (* Alexander Adamchuk, Jun 30 2006 *)
RecurrenceTable[{a[1]==1,a[n]==n^n-a[n-1]},a,{n,20}] (* Harvey P. Dale, Jan 21 2015 *)

from itertools import accumulate, count, islice
def A001099_gen(): # generator of terms
    yield from accumulate((k**k for k in count(1)),func=lambda x,y:y-x)
A001099_list = list(islice(A001099_gen(),20)) # Chai Wah Wu, Jun 17 2022

Absolute value of Sum_{k=1..n} k^k*(-1)^(k+1). a(n) = n^n - (n-1)^(n-1) + (n-2)^(n-2) - ... - (-1)^n*1^1. - Alexander Adamchuk, Jun 30 2006

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User: Ed Smiley

A226502 Let P(k) denote the k-th prime (P(1)=2, P(2)=3 ...); a(n) = P(n+1)P(n+3) - P(n)P(n+2).

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A182664 a(n) = A088828(n) + A157502(n).

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A001099 a(n) = n^n - a(n-1), with a(1) = 1.

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Python

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