A016945 -id:A016945 - cp's OEIS frontend

A188378 Partial sums of A005248.

2, 5, 12, 30, 77, 200, 522, 1365, 3572, 9350, 24477, 64080, 167762, 439205, 1149852, 3010350, 7881197, 20633240, 54018522, 141422325, 370248452, 969323030, 2537720637, 6643838880, 17393796002, 45537549125, 119218851372, 312119004990, 817138163597, 2139295485800

Offset: 0

Gabriele Fici, Mar 29 2011

Different from A024851.

Luo proves that these integers cannot be uniquely decomposed as the sum of distinct and nonconsecutive terms of the Lucas number sequence. - Michel Marcus, Apr 20 2020

Robert Israel, Table of n, a(n) for n = 0..1000
Clark Kimberling, Lucas Representations of Positive Integers, J. Int. Seq., Vol. 23 (2020), Article 20.9.5.
David C. Luo, Nonuniqueness Properties of Zeckendorf Related Decompositions, arXiv:2004.08316 [math.NT], 2020.
Index entries for linear recurrences with constant coefficients, signature (4,-4,1).

Cf. A000032, A002878, A027941, A001654.

Cf. A016945, A267797.

[5*Fibonacci(n)*Fibonacci(n+1)+1+(-1)^n: n in [0..40]]; // Vincenzo Librandi, Jan 24 2016

f:= gfun:-rectoproc({a(n+3)-4*a(n+2)+4*a(n+1)-a(n), a(0) = 2, a(1) = 5, a(2) = 12}, a(n), remember):
map(f, [$0..60]); # Robert Israel, Feb 02 2016

LinearRecurrence[{4,-4,1},{2,5,12},30] (* Harvey P. Dale, Oct 05 2015 *)
Accumulate@ LucasL@ Range[0, 58, 2] (* Michael De Vlieger, Jan 24 2016 *)

a(n) = 5*fibonacci(n)*fibonacci(n+1) + 1 + (-1)^n; \\ Michel Marcus, Aug 26 2013

Vec((-2+3*x)/((x-1)*(x^2-3*x+1)) + O(x^100)) \\ Altug Alkan, Jan 24 2016

a(n) = A000032(2n+1)+1 = A002878(n)+1 = 2*A027941(n+1)-3*A027941(n).
G.f.: ( -2+3*x ) / ( (x-1)*(x^2-3*x+1) ). - R. J. Mathar, Mar 30 2011
a(n) = 5*A001654(n) + 1 + (-1)^n, n>=0. [Wolfdieter Lang, Jul 23 2012]
(a(n)^3 + (a(n)-2)^3) / 2 = A000032(A016945(n)) = Lucas(6*n+3) = A267797(n), for n>0. - Altug Alkan, Jan 31 2016
a(n) = 2^(-1-n)*(2^(1+n)-(3-sqrt(5))^n*(-1+sqrt(5))+(1+sqrt(5))*(3+sqrt(5))^n). - Colin Barker, Nov 02 2016

A271209 a(n) = n^5 + n + 1.

Original entry on oeis.org

1, 3, 35, 247, 1029, 3131, 7783, 16815, 32777, 59059, 100011, 161063, 248845, 371307, 537839, 759391, 1048593, 1419875, 1889587, 2476119, 3200021, 4084123, 5153655, 6436367, 7962649, 9765651, 11881403, 14348935, 17210397, 20511179, 24300031, 28629183, 33554465

Offset: 0

Jaroslav Krizek, Apr 02 2016

For n>1 these are odd composite numbers: all terms a(n) are divisible by number h(n) = GCD(n^5+n+1,(n+1)^5+n) = GCD(a(n), a(n+1)-2) = (n*(n+1)+1)*GCD(n*(n+1)-1, 5) where 1 < h(n) < a(n) for all n>1. Sequence of corresponding numbers h(n) for n>1: 35, 13, 21, 31, 43, 285, ... For example, a(7) = 16815 is divisible by number h(7) = (7*(7+1)+1)*GCD(7*(7+1)-1, 5) = 57*GCD(55, 5) = 57*5 = 285.
We name a set of k sequences IOPR_k(n) = {a_1(n) = a(n), a_2(n) =  a(n) + 2, ...,  a_k(n) = a(n) + 2*(k - 1)} as infinite nonprime k-lane road if a arithmetic function a(n) defined by arithmetic operations produces for all n > h (h = a small integer >= 0) odd terms such that all values a(n), a(n) + 2, ..., a(n) + 2*(k - 1) are composites. We say sequences a_1(n) = a(n), a_2(n) =  a(n) + 2, ...,  a_k(n) = a(n) + 2*(k - 1) are k-th lanes of set IOPR_k(n).
For example, sequence A016945(n) = 6*n + 3 = IOPR_1(n) for k=1.
This sequence a(n) is 2nd lane of set of sequences IOPR_2(n) = {a_1(n) = A271208(n) = a(n) - 2 = n^5 + n - 1, a_2(n) = a(n) = n^5 + n + 1}.
If p = prime > 2 of the form 3m - 1 from A003627 then sets of 2 sequences {n^p + n - 1,  n^p + n + 1} = IOPR_2(n) for all p.
Also sets of 2 sequences {n^k + n - 1,  n^k + n + 1} = IOPR_2(n) for all k>2 from A016789.
In general, if k>2 is number of the form 3m - 1 from A016789 then sequences a(n) = n^k + n - 1 and  b(n) = a(n) + 2 = n^k + n + 1 produces for all n > 1 odd composite terms. The terms of sequence a(n) = n^k + n - 1 are divisible for all n > 1 by number h(n) = GCD(n^k+n-1,(n-1)^k+n) = GCD(a(n), a(n-1)+2) = (n*(n-1)+1)*GCD(n*(n-1)-1, k) where 1 < h(n) < a(n) for all n>1. The terms of sequence b(n) = a(n) + 2 = n^k + n + 1 are divisible for all n > 1 by number h(n) = GCD(n^k+n+1,(n+1)^k+n) = GCD(a(n), a(n+1)-2) = (n*(n+1)+1)*GCD(n*(n+1)-1, k)  where 1 < h(n) < a(n) for all n>1.
Are there any sets of sequences IOPR_k(n) for k>2? For example, like set of sequences {A161945(n), A161945(n) + 2, A161945(n) + 4} is not an infinite nonprime 3-lane road because sequence A161945 is not defined by arithmetic operations.

Index entries for linear recurrences with constant coefficients, signature (6,-15,20,-15,6,-1).

Cf. A003627, A016789, A131471, A161945, A271208.

Magma
```
[n^5+n+1: n in[0..100]];
```

A271209:=n->n^5 + n + 1: seq(A271209(n), n=0..40); # Wesley Ivan Hurt, Apr 02 2016

Table[n^5+n+1, {n, 0, 100}] (* Waldemar Puszkarz, Apr 02 2016 *)
LinearRecurrence[{6,-15,20,-15,6,-1},{1,3,35,247,1029,3131},40] (* Harvey P. Dale, Jul 24 2016 *)

for(n=0, 100, print1(n^5+n+1, ", ")) \\ Waldemar Puszkarz, Apr 02 2016

a(n) = A271208(n) + 2.
From Wesley Ivan Hurt, Apr 02 2016: (Start)
G.f.: (1-3*x+32*x^2+62*x^3+27*x^4+x^5) / (x-1)^6.
a(n) = 6*a(n-1)-15*a(n-2)+20*a(n-3)-15*a(n-4)+6*a(n-5)-a(n-6), n>5. (End)
a(n) = A131471(n) + 1. - Omar E. Pol, Apr 05 2016

A332075 Odd numbers 2n+1 such that k + 2^m is prime, where k and m are the odd part and 2-valuation, respectively, of 2n.

Original entry on oeis.org

3, 5, 7, 11, 13, 17, 19, 23, 25, 29, 31, 35, 37, 41, 43, 49, 53, 55, 59, 61, 71, 73, 77, 79, 83, 89, 91, 101, 103, 109, 113, 115, 119, 121, 131, 133, 139, 143, 149, 155, 157, 161, 163, 169, 173, 175, 185, 191, 193, 197, 199, 203, 209, 211, 215, 221, 223, 229, 233, 241, 251

Offset: 1

M. F. Hasler, Aug 13 2020

nonn

It appears that about 1/log_10(N) of the odd numbers below 2N are in this sequence: for n < 10^k with k = (1, 2, 3, 4, 5, 6), there are (7, 51, 364, 2675, 20668, 167185) numbers as defined in NAME.
We observe that the sequence mainly consists of the odd primes but some of them are missing (47, 67, 97, 107, 127, 137, 151, 167, ...) and there are some composite terms {25, 35, 49, 55, 77, 91, ...}.
The frequency of primes in this sequence remains high: the least prime > 10^99 with this property is only 10^99 + 2191. See A332078 for primes not in this sequence.
These are numbers of the form (p-2^m)*2^m + 1, where p is an odd prime and 1 < 2^m < p, so there are infinitely many such numbers. Problem: are there infinitely many primes of this form? All the numbers A016945 > 3 are not in this sequence. - Thomas Ordowski, Aug 13 2020
Otherwise said, 3 is the only term divisible by 3. - M. F. Hasler, Aug 14 2020

			For the smallest odd number 1 = 2*0 + 1, 2n = 2*0 has no well defined 2-adic valuation, so it is omitted here.
For the next odd number 3 = 2*1 + 1, 2n = 2 has odd part (A000265) and 2-adic valuation (A007814) both equal to k = m = 1, and k + 2^m = 3 is prime.

Michel Marcus, Table of n, a(n) for n = 1..6000
Thomas Ordowski, Problem, post to the SeqFan list, Aug 11 2020.

Cf. A000040 (primes), A000265 (odd part), A007814 (2-valuation), A332078 (primes that are not in this sequence), A332079 (number of primes between 2^n and the next larger prime in this sequence).

Select[Range[3, 251, 2], PrimeQ[(m = 2^IntegerExponent[# - 1, 2]) + (# - 1)/m] &] (* Amiram Eldar, Aug 14 2020 *)

select( is_A332075(n)=n%2&&n>1&&ispseudoprime((n>>n=valuation(n-1,2))+2^n), [1..255])

A337940 Triangle read by rows: T(n, k) = T(n+2) - T(n-k), with the triangular numbers T = A000217, for n >= 1, k = 1, 2, ..., n.

Original entry on oeis.org

6, 9, 10, 12, 14, 15, 15, 18, 20, 21, 18, 22, 25, 27, 28, 21, 26, 30, 33, 35, 36, 24, 30, 35, 39, 42, 44, 45, 27, 34, 40, 45, 49, 52, 54, 55, 30, 38, 45, 51, 56, 60, 63, 65, 66, 33, 42, 50, 57, 63, 68, 72, 75, 77, 78, 36, 46, 55, 63, 70, 76, 81, 85, 88, 90, 91

Offset: 1

Wolfdieter Lang, Nov 23 2020

This number triangle results from the array A(n, m) = T(n+m+1) - T(n-1), with T = A000217, for n, m >= 1. For this array see the example by Bob Selcoe, in A111774 (but with rows continued). The present triangle is obtained by reading the array by upwards antidiagonals: T(n, k) = A(n+1-k, k). See also the Jul 09 2019 comment by Ralf Steiner with the formula c_k(n) (rows k >= 1, columns n >= 3), rewritten for A(n, m) = (m+2)*(2*n+m+1)/2, leading to T(n, k) = (k+2)*(2*n-k+3)/2.
Therefore this triangle is related to the problem of giving the numbers which are sums of at least three consecutive positive integers given as sequence A111774. It allows us to find the multiplicities for the numbers of A111774. They are given in A338428(n).
To obtain the multiplicity for number N (>= 6) from A111774 one has to consider only the triangle rows n = 1, 2, ..., floor((N-3)/3).
The row reversed triangle, considered by Bob Selcoe in A111774, is T(n, n-k+1) = T(n+2) - T(k-1), for n >= 1, and k=1, 2, ..., n.
This triangle contains no odd prime numbers and no exact powers 2^m, for m >= 0. This can be seen by considering the diagonal sequences D(d, k), for d >= 1, k >= 1 or the row sequences of the array A(n, m), for n >= 1 and m >= 1. The result is A(r+1, s-2) = s*(s + 2*r + 1)/2, for r >= 0 and s >= 3 (from the g.f. of the diagonals of T given below). This is also given in the Jul 09 2019 comment by Ralf Steiner in A111774. Therefore A(r+1, s-2) is a product of two numbers >= 2, hence not a prime. And in both cases (i) s/2 integer or (ii) (s + 2*r + 1)/2 integer not both numbers can be powers of 2 by simple parity arguments.
The previous comment means that each T(n, k) has at least one odd prime as a proper divisor.
A number N appears in this triangle, or in A111774, if and only if floor(N/2) - delta(N) >= 1, where delta(N) = A055034(N). For the sequence b(n) := floor(n/2) - delta(n), for n >= 2, see A219839(n), b(1) = -1. See a W. Lang comment in A111774 for the proof.

			The triangle T(n, k) begins:
n \ k  1  2  3  4  5   6   7   8   9  10  11  12  13  14  15 ...
1:     6
2:     9 10
3:    12 14 15
4:    15 18 20 21
5:    18 22 25 27 28
6:    21 26 30 33 35  36
7:    24 30 35 39 42  44  45
8:    27 34 40 45 49  52  54  55
9:    30 38 45 51 56  60  63  65  66
10:   33 42 50 57 63  68  72  75  77  78
11:   36 46 55 63 70  76  81  85  88  90  91
12:   39 50 60 69 77  84  90  95  99 102 104 105
13:   42 54 65 75 84  92  99 105 110 114 117 119 120
14:   45 58 70 81 91 100 108 115 121 126 130 133 135 136
15:   48 62 75 87 98 108 117 125 132 138 143 147 150 152 153
...
N = 15 appears precisely twice from the sums 4+5+6 = A(4, 1) = T(4, 1), and (1+2+3)+4+5 = A(1, 3) = T(3, 3), i.e., with a sum of 3 and 5 consecutive positive integers.
N = 42 appears three times from the sums 13+14+15 = A(13, 1) = T(13, 1), 9+10+11 +12 = A(9, 2) = T(10, 2), 3+4+5+6+7+8+9 = A(3, 5) = T(7, 5); i.e., 42 can be written as a sum of 3, 4 and 7 consecutive positive integers.

Cf. A055034, A111774, A338428 (multiplicities), A219839.
For columns k = 1, 2, ..., 10 see A008585, A016825, A008587, A016945, A008589, A017113, A008591, A017329, A008593, A017593.
For diagonals d = 1, 2, ..., 10 see A000217, A000096, A055998, A055999, A056000, A056115, A056119 , A056121, A056126, A051942.

Flatten[Table[((n+2)*(n+3)-(n-k)*(n-k+1))/2,{n,11},{k,n}]] (* Stefano Spezia, Nov 24 2020 *)

T(n, k) = ((n+2)*(n+3) - (n-k)*(n-k+1))/2, for n >= 1 and k = 1, 2, ..., n (see the name).
T(n, k) = (k+2)*(2*n-k+3)/2 (factorized).
G.f. columns k = 2*j+1, for j >= 0: Go(j, x) = x^(2*j+1)*(2*j+3)*(j+2 - (j+1)*x)/(1-x)^2,
G.f. columns k = 2*j, for j >= 1: Ge(j, x) = x^(2*j)*(j+1)*(2*j+3 - (2*j+1)*x)/(1-x)^2.
G.f. row polynomials: G(z,x) = z*x*(1 + z*x)^3*{3*(2-z) - (8-3*z)*(z*x) + (3-z)*(z*x)^2}/((1 - z)^2*(1 - (z*x)^2)^3).
G.f. diagonals d >= 1: GD(d, x) = ((d+1)*3 - (5*d+3)*x + (2*d+1)*x^2)/(1-x)^3.
G.f. of GD(d, x): GGD(z,x) = (6-8*x+3*x^2 - (3-3*x+x^2)*z)/((1-x)^3*(1-z)^2).

A359376 Numbers that are either odd multiples of 3 or multiples of 4. Numbers k such that A252463(k) is even.

Original entry on oeis.org

0, 3, 4, 8, 9, 12, 15, 16, 20, 21, 24, 27, 28, 32, 33, 36, 39, 40, 44, 45, 48, 51, 52, 56, 57, 60, 63, 64, 68, 69, 72, 75, 76, 80, 81, 84, 87, 88, 92, 93, 96, 99, 100, 104, 105, 108, 111, 112, 116, 117, 120, 123, 124, 128, 129, 132, 135, 136, 140, 141, 144, 147, 148, 152, 153, 156, 159, 160, 164, 165

Offset: 1

Antti Karttunen, Dec 31 2022

Numbers that are congruent to {0, 3, 4, 8, 9} mod 12. - Amiram Eldar, Jan 24 2023

Amiram Eldar, Table of n, a(n) for n = 1..10000
Index entries for linear recurrences with constant coefficients, signature (1,0,0,0,1,-1).

Disjoint union of A016945 and A008586.
Cf. A000035, A064989, A252463, A359375 (complement), A359379 (characteristic function).
Positions of 0's in A359374.

Select[Range[0, 165], Mod[#, 6] == 3 || Mod[#, 4] == 0 &] (* Amiram Eldar, Jan 24 2023 *)

PARI
```
isA359376(n) = A359379(n);
```
PARI
```
isA359376(n) = !A359374(n);
```

isA359376(n) = (((n%2)&&!(n%3))||!(n%4));

A016954 a(n) = (6n+3)^10.

Original entry on oeis.org

59049, 3486784401, 576650390625, 16679880978201, 205891132094649, 1531578985264449, 8140406085191601, 34050628916015625, 119042423827613001, 362033331456891249, 984930291881790849, 2446194060654759801, 5631351470947265625, 12157665459056928801

Offset: 0

N. J. A. Sloane

Vincenzo Librandi, Table of n, a(n) for n = 0..2000
Index entries for linear recurrences with constant coefficients, signature (11,-55,165,-330,462,-462,330,-165,55,-11,1).

Cf. A008454, A016762, A016945, A016946, A016947, A016948, A016949, A016950, A016951, A016952, A016953.

Subsequence of A008454.

[(6*n+3)^10: n in [0..20]]; // Vincenzo Librandi, May 06 2011

A016954:=n->(6*n+3)^10: seq(A016954(n), n=0..20); # Wesley Ivan Hurt, Aug 22 2016

(6 Range[0, 20] + 3)^10 (* Wesley Ivan Hurt, Aug 22 2016 *)

From Wesley Ivan Hurt, Aug 22 2016: (Start)
G.f.: 59049*(1 + 59038*x + 9116141*x^2 + 178300904*x^3 + 906923282*x^4 + 1527092468*x^5 + 906923282*x^6 + 178300904*x^7 + 9116141*x^8 + 59038*x^9 + x^10)/(1-x)^11.
a(n) = 11*a(n-1) - 55*a(n-2) + 165*a(n-3) - 330*a(n-4) + 462*a(n-5) - 462*a(n-6) + 330*a(n-7) - 165*a(n-8) + 55*a(n-9) - 11*a(n-10) + a(n-11) for n>10.
a(n) = A008454(A016945(n)). (End)
From Amiram Eldar, Mar 30 2022: (Start)
a(n) = A016946(n)^5 = A016949(n)^2.
a(n) = 3^10*A016762(n).
Sum_{n>=0} 1/a(n) = 31*Pi^10/171421608960. (End)

A124011 Add three, add six, add nine, ....

Original entry on oeis.org

5, 8, 14, 23, 35, 50, 68, 89, 113, 140, 170, 203, 239, 278, 320, 365, 413, 464, 518, 575, 635, 698, 764, 833, 905, 980, 1058, 1139, 1223, 1310, 1400, 1493, 1589, 1688, 1790, 1895, 2003, 2114, 2228, 2345, 2465, 2588, 2714, 2843, 2975, 3110, 3248, 3389, 3533, 3680

Offset: 1

Ruben Fritzon (ruben.fritzon(AT)edu.ovanaker.se), Dec 11 2006

Found on a quiz.

Vincenzo Librandi, Table of n, a(n) for n = 1..1000
Index entries for linear recurrences with constant coefficients, signature (3,-3,1).

Cf. A000217, A016945, A045943.

I:=[5,8,14]; [n le 3 select I[n] else 3*Self(n-1)-3*Self(n-2)+1*Self(n-3): n in [1..50]]; // Vincenzo Librandi, Feb 15 2012

LinearRecurrence[{3,-3,1},{5,8,14},50] (* Vincenzo Librandi, Feb 15 2012 *)
Table[(3x^2-3x+10)/2,{x,50}] (* Harvey P. Dale, Jul 25 2019 *)
Accumulate[3*Range[0,50]]+5 (* Harvey P. Dale, Jan 16 2024 *)

a(n)=3*n*(n-1)/2+5 \\ Charles R Greathouse IV, Jun 17 2017

a(n) = 3*n*(n-1)/2 + 5 = 3*A000217(n-1) + 5 = A045943(n) + 5.
a(n) = 3*n + a(n-1) - 3 with a(1)=5. - Vincenzo Librandi, Nov 28 2009
G.f.: x*(5 - 7*x + 5*x^2)/(1-x)^3. - Colin Barker, Jan 14 2012
3*(8*a(n) - 37) = A016945(n-1)^2. - Vincenzo Librandi, Feb 15 2012
From Elmo R. Oliveira, Feb 11 2025: (Start)
E.g.f.: exp(x)*(5 + 3*x^2/2) - 5.
a(n) = 3*a(n-1) - 3*a(n-2) + a(n-3) for n > 3. (End)

More terms from Graham Roy (groy(AT)ashland.edu), Dec 14 2006

Additional comments from Christopher N. Swanson (cswanson(AT)ashland.edu), R. J. Mathar, Dec 14 2006

A139482 Binomial transform of [1, 1, 2, 1, -1, 1, -1, 1, -1, 1, -1, 1, -1, ...].

Original entry on oeis.org

1, 2, 5, 11, 20, 32, 47, 65, 86, 110, 137, 167, 200, 236, 275, 317, 362, 410, 461, 515, 572, 632, 695, 761, 830, 902, 977, 1055, 1136, 1220, 1307, 1397, 1490, 1586, 1685, 1787, 1892, 2000, 2111, 2225

Offset: 1

Gary W. Adamson, Apr 23 2008

A007318 * [1, 1, 2, 1, -1, 1, -1, 1, ...].
The quadratic expression for a(n) follows at once by taking into account that the alternate row sums in the Pascal triangle are equal to zero (starting with the second row). - Emeric Deutsch, May 03 2008
For n > 1, 3*(8*a(n) - 13) = A016945(n-2)^2. - Vincenzo Librandi, Feb 15 2012

			a(4) = 11 = (1, 3, 3, 1) dot (1, 1, 2, 1) = (1 + 3 + 6 + 1).

Vincenzo Librandi, Table of n, a(n) for n = 1..1000
Index entries for linear recurrences with constant coefficients, signature (3,-3,1).

1,seq((10+3*n^2-9*n)*1/2,n=2..40); # Emeric Deutsch, May 03 2008

Join[{1,2},FoldList[##+3&,5,3*Range@100]] (* Vladimir Joseph Stephan Orlovsky, Feb 17 2011 *)
LinearRecurrence[{3,-3,1},{1,2,5,11},40] (* Harvey P. Dale, May 02 2015 *)

G.f.: (x^3+2*x^2-x+1)/(-x^3+3*x^2-3*x+1). - Alexander R. Povolotsky, Apr 24 2008
a(n) = (10 - 9*n + 3*n^2)/2 for n >= 2. - Emeric Deutsch, May 03 2008
a(n) = 3*a(n-1) - 3*a(n-2) + a(n-3); a(1)=1, a(2)=2, a(3)=5, a(4)=11. - Harvey P. Dale, May 02 2015

More terms from Emeric Deutsch, May 03 2008

A157932 Numbers k such that (3^(35k) + 5^(21k) + 7^(15*k)) mod 105 is prime.

Original entry on oeis.org

0, 4, 6, 8, 12, 16, 18, 20, 24, 28, 30, 32, 36, 40, 42, 44, 48, 52, 54, 56, 60, 64, 66, 68, 72, 76, 78, 80, 84, 88, 90, 92, 96, 100, 102, 104, 108, 112, 114, 116, 120, 124, 126, 128, 132, 136, 138, 140, 144, 148, 150, 152, 156, 160, 162, 164, 168, 172, 174

Offset: 1

Kyle D. Balliet, Mar 09 2009

Let b(k) = (3^(35*k) + 5^(21*k) + 7^(15*k)) mod 105, then sequence {b(k)} is 3, repeat (60, 68, 75, 17, 30, 23, 60, 47, 75, 38, 30, 2), with primes 3, 17, 23, 47, 2. First differences of {a(n)} are 4, 2, 2, 4, 4, 2, 2, 4, .... - Michel Marcus, Aug 15 2013
3^(35*k) + 5^(21*k) + 7^(15*k) = (4^k)*(3^k + 5^k + 7^k) mod 105, then by the division algorithm a simple proof yields that only numbers k of the form 24*m, 24*m+4, 24*m+6, 24*m+8, 24*m+12, 24*m+16, 24*m+18, 24*m+20 will be congruent to a prime modulo 105. Thus the pattern 4, 2, 2, 4, 4, 2, 2, ... will repeat infinitely. - Kyle D. Balliet, Jan 01 2014
Even numbers that can be written as the sum of 3 of their divisors, not necessarily distinct (see A355200). Also, numbers k of the form 12*m, 12*m+4, 12*m+6, 12*m+8. - Bernard Schott, Sep 08 2023

			a(4) = 3^(35*4) + 5^(21*4) + 7^(15*4) mod 105 = 17 (prime).

Colin Barker, Table of n, a(n) for n = 1..1000
Index entries for linear recurrences with constant coefficients, signature (2,-2,2,-1).

Equals {0} Union (A355200 \ A016945) <=> subsequence of even terms in A355200.

Select[Range[0,180],PrimeQ[Mod[3^(35#)+5^(21#)+7^(15#),105]]&] (* Harvey P. Dale, Oct 10 2017 *)

isok(n) = isprime((3^(35*n)+5^(21*n)+7^(15*n)) % 105); \\ Michel Marcus, Aug 15 2013

a(n)=n\4*12+[-4,0,4,6][n%4+1] \\ Charles R Greathouse IV, Dec 27 2013

is(n)=n%=12;n==0||n==4||n==6||n==8 \\ Charles R Greathouse IV, Dec 27 2013

a(n) = (-6-(-I)^n-I^n+6*n)/2 \\ Colin Barker, Oct 19 2015

concat(0, Vec(2*x^2*(2*x^2-x+2)/((x-1)^2*(x^2+1)) + O(x^100))) \\ Colin Barker, Oct 19 2015

3n - 4 <= a(n) <= 3n - 2. - Charles R Greathouse IV, Dec 27 2013
From Colin Barker, Oct 19 2015: (Start)
a(n) = (-6 - (-i)^n - i^n + 6*n)/2, where i = sqrt(-1).
G.f.: 2*x^2*(2*x^2-x+2) / ((x-1)^2*(x^2+1)). (End)

More terms from Michel Marcus, Aug 15 2013

A194801 Square array read by antidiagonals: T(n,k) = k((n+1)k-n+1)/2, k = 0, +- 1, +- 2,..., n >= 0.

Original entry on oeis.org

0, 0, 1, 0, 1, 0, 0, 1, 1, 3, 0, 1, 2, 4, 1, 0, 1, 3, 5, 4, 6, 0, 1, 4, 6, 7, 9, 3, 0, 1, 5, 7, 10, 12, 9, 10, 0, 1, 6, 8, 13, 15, 15, 16, 6, 0, 1, 7, 9, 16, 18, 21, 22, 16, 15, 0, 1, 8, 10, 19, 21, 27, 28, 26, 25, 10, 0, 1, 9, 11, 22, 24, 33, 34, 36, 35

Offset: 0

Omar E. Pol, Feb 05 2012

Note that a single formula gives several types of numbers. Row 0 lists 0 together the Molien series for 3-dimensional group [2,k]+ = 22k. Row 1 lists, except first zero, the squares repeated. If n >= 2, row n lists the generalized (n+3)-gonal numbers, for example: row 2 lists the generalized pentagonal numbers A001318. See some other examples in the cross-references section.

			Array begins:
(A008795): 0, 1,  0,  3,  1,  6,  3, 10,   6,  15,  10...
(A008794): 0, 1,  1,  4,  4,  9,  9, 16,  16,  25,  25...
A001318:   0, 1,  2,  5,  7, 12, 15, 22,  26,  35,  40...
A000217:   0, 1,  3,  6, 10, 15, 21, 28,  36,  45,  55...
A085787:   0, 1,  4,  7, 13, 18, 27, 34,  46,  55,  70...
A001082:   0, 1,  5,  8, 16, 21, 33, 40,  56,  65,  85...
A118277:   0, 1,  6,  9, 19, 24, 39, 46,  66,  75, 100...
A074377:   0, 1,  7, 10, 22, 27, 45, 52,  76,  85, 115...
A195160:   0, 1,  8, 11, 25, 30, 51, 58,  86,  95, 130...
A195162:   0, 1,  9, 12, 28, 33, 57, 64,  96, 105, 145...
A195313:   0, 1, 10, 13, 31, 36, 63, 70, 106, 115, 160...
A195818:   0, 1, 11, 14, 34, 39, 69, 76, 116, 125, 175...

Rows (0-11): 0 together with A008795, (truncated A008794), A001318, A000217, A085787, A001082, A118277, A074377, A195160, A195162, A195313, A195818
Columns (0-9): A000004, A000012, A001477, (truncated A000027), A016777, (truncated A008585), A016945, (truncated A016957), A017341, (truncated A017329).
Cf. A139600.

cp's OEIS Frontend

A188378 Partial sums of A005248.

Views

Author

Keywords

Comments

Links

Crossrefs

Programs

Magma

Maple

Mathematica

PARI

PARI

Formula

A271209 a(n) = n^5 + n + 1.

Views

Author

Keywords

Comments

Links

Crossrefs

Programs

Magma

Maple

Mathematica

PARI

Formula

A332075 Odd numbers 2n+1 such that k + 2^m is prime, where k and m are the odd part and 2-valuation, respectively, of 2n.

Views

Author

Keywords

Comments

Examples

Links

Crossrefs

Programs

Mathematica

PARI

A337940 Triangle read by rows: T(n, k) = T(n+2) - T(n-k), with the triangular numbers T = A000217, for n >= 1, k = 1, 2, ..., n.

Views

Author

Keywords

Comments

Examples

Crossrefs

Programs

Mathematica

Formula

A359376 Numbers that are either odd multiples of 3 or multiples of 4. Numbers k such that A252463(k) is even.

Views

Author

Keywords

Comments

Links

Crossrefs

Programs

Mathematica

PARI

PARI

PARI

A016954 a(n) = (6n+3)^10.

Views

Author

Keywords

Links

Crossrefs

Programs

Magma

Maple

Mathematica

Formula

A124011 Add three, add six, add nine, ....

Views

Author

Keywords

Comments

Links

Crossrefs

Programs

A157932 Numbers k such that (3^(35k) + 5^(21k) + 7^(15*k)) mod 105 is prime.

A194801 Square array read by antidiagonals: T(n,k) = k((n+1)k-n+1)/2, k = 0, +- 1, +- 2,..., n >= 0.