W. Neville Holmes - cp's OEIS frontend

A130699 Numbers n for which neither 2n-3 nor 2n+3 are primes.

6, 9, 12, 15, 18, 21, 24, 26, 27, 30, 33, 36, 39, 42, 44, 45, 48, 51, 54, 57, 59, 60, 61, 63, 66, 69, 72, 75, 78, 79, 81, 84, 86, 87, 90, 93, 96, 99, 102, 103, 105, 106, 108, 109, 111, 114, 117, 120, 123, 125, 126, 128, 129, 131, 132, 135, 138, 141, 144, 146

Offset: 2

W. Neville Holmes, Jul 11 2007

			Not 5 because 7 and 13 are prime, but 6 because neither 9 nor 15 are primes.

Cf. A104278.

Select[Range[200],NoneTrue[2#+{3,-3},PrimeQ]&] (* The program uses the NoneTrue function from Mathematica version 10 *) (* Harvey P. Dale, Nov 12 2014 *)

isok(n) = !isprime(2*n-3) && !isprime(2*n+3) \\ Michel Marcus, Jul 11 2013

Missing term 24 added by Michel Marcus, Jul 11 2013

A130771 Numbers n not divisible by 3 for which neither 2n-3 nor 2n+3 are primes.

Original entry on oeis.org

26, 44, 59, 61, 79, 86, 103, 106, 109, 125, 128, 131, 146, 149, 151, 161, 163, 166, 169, 179, 184, 187, 194, 205, 224, 236, 239, 254, 257, 265, 266, 268, 271, 274, 278, 281, 289, 293, 296, 304, 313, 316, 326, 334, 341, 346, 350, 355, 359, 364, 367, 376, 389, 391

Offset: 1

W. Neville Holmes, Jul 14 2007

This is A130699 with the trivial multiples of 3 removed.

			26 because 2*26+3=55 is not a prime, nor is 2*26-3=49.

Harvey P. Dale, Table of n, a(n) for n = 1..2500

Cf. A130699, A001651.

Select[Range[400],!Divisible[#,3]&&NoneTrue[2#+{3,-3},PrimeQ]&] (* Requires Mathematica version 10 or later *) (* Harvey P. Dale, Sep 19 2018 *)

isok(n) = (n % 3) && !isprime(2*n-3) && !isprime(2*n+3) \\ Michel Marcus, Jul 11 2013

Missing term 169 added by Michel Marcus, Jul 11 2013

A130723 Least common multiple of 3 and n^2+n+1.

Original entry on oeis.org

3, 3, 21, 39, 21, 93, 129, 57, 219, 273, 111, 399, 471, 183, 633, 723, 273, 921, 1029, 381, 1263, 1389, 507, 1659, 1803, 651, 2109, 2271, 813, 2613, 2793, 993, 3171, 3369, 1191, 3783, 3999, 1407, 4449, 4683, 1641, 5169, 5421, 1893, 5943, 6213, 2163, 6771

Offset: 0

W. Neville Holmes, Jul 04 2007

Harvey P. Dale, Table of n, a(n) for n = 0..1000
Index entries for linear recurrences with constant coefficients, signature (0,0,3,0,0,-3,0,0,1).

Cf. A109044.

Table[LCM[3,n^2+n+1],{n,0,60}] (* Harvey P. Dale, Mar 03 2017 *)

a(n) = lcm(3, n^2+n+1) \\ Michel Marcus, Jul 11 2013

Vec(3*(1 + x + 7*x^2 + 10*x^3 + 4*x^4 + 10*x^5 + 7*x^6 + x^7 + x^8) / ((1 - x)^3*(1 + x + x^2)^3) + O(x^100)) \\ Colin Barker, Mar 08 2017

From Colin Barker, Mar 08 2017: (Start)
G.f.: 3*(1 + x + 7*x^2 + 10*x^3 + 4*x^4 + 10*x^5 + 7*x^6 + x^7 + x^8) / ((1 - x)^3*(1 + x + x^2)^3).
a(n) = 3*a(n-3) - 3*a(n-6) + a(n-9) for n>8.
(End)

Corrected by Harvey P. Dale, Mar 03 2017

A130770 One third of the least common multiple of 3 and n^2+n+1.

Original entry on oeis.org

1, 1, 7, 13, 7, 31, 43, 19, 73, 91, 37, 133, 157, 61, 211, 241, 91, 307, 343, 127, 421, 463, 169, 553, 601, 217, 703, 757, 271, 871, 931, 331, 1057, 1123, 397, 1261, 1333, 469, 1483, 1561, 547, 1723, 1807, 631, 1981, 2071, 721, 2257, 2353, 817, 2551, 2653, 919

Offset: 0

W. Neville Holmes, Jul 14 2007

This is a subset of A051176 and is also one third of A130723.

			a(4)=7 because 4^2+4+1 =21, the LCM of 3 and 21 is 21 and 21/3=7.

G. C. Greubel, Table of n, a(n) for n = 0..5000
Index entries for linear recurrences with constant coefficients, signature (0, 0, 3, 0, 0, -3, 0, 0, 1).

Cf. A051176, A130723.

[Lcm(3,n^2+n+1)/3: n in [0..50]]; // G. C. Greubel, Oct 26 2017

seq(denom((n-1)^2/(n^2+n+1)), n=0..52) ; # Zerinvary Lajos, Jun 04 2008

Table[LCM[3,n^2+n+1]/3,{n,0,60}] (* or *) LinearRecurrence[ {0,0,3,0,0,-3,0,0,1},{1,1,7,13,7,31,43,19,73},60] (* Harvey P. Dale, Apr 10 2014 *)

for(n=0,50, print1(lcm(3, n^2 + n +1)/3, ", ")) \\ G. C. Greubel, Oct 26 2017

Conjecture: a(n) = A046163(n), n>0. - R. J. Mathar, Jun 13 2008

a(n) = 3*a(n-3)-3*a(n-6)+a(n-9), with a(0)=1, a(1)=1, a(2)=7, a(3)=13, a(4)=7, a(5)=31, a(6)=43, a(7)=19, a(8)=73. - Harvey P. Dale, Apr 10 2014

A130724 a(n) = lcm(n,3) / gcd(n,3).

Original entry on oeis.org

0, 3, 6, 1, 12, 15, 2, 21, 24, 3, 30, 33, 4, 39, 42, 5, 48, 51, 6, 57, 60, 7, 66, 69, 8, 75, 78, 9, 84, 87, 10, 93, 96, 11, 102, 105, 12, 111, 114, 13, 120, 123, 14, 129, 132, 15, 138, 141, 16, 147, 150, 17, 156, 159, 18, 165, 168, 19, 174, 177, 20, 183, 186

Offset: 0

W. Neville Holmes, Jul 04 2007

			a(7) = 21 because lcm(3,7) = 21, gcd(3,7) = 1 and 21/1 = 21.

Index entries for linear recurrences with constant coefficients, signature (0,0,2,0,0,-1).

Cf. A109007, A109044.

[Lcm(n,3)/Gcd(n,3) : n in [0..100]]; // Wesley Ivan Hurt, Jul 24 2016

A130724:=n->lcm(n,3)/gcd(n,3): seq(A130724(n), n=0..100); # Wesley Ivan Hurt, Jul 24 2016

LCM[3,#]/GCD[3,#]&/@Range[0,70] (* Harvey P. Dale, May 16 2013 *)

a(n) = A109044(n) / A109007(n).
From Wesley Ivan Hurt, Jul 24 2016: (Start)
G.f.: x*(3 + 6*x + x^2 + 6*x^3 + 3*x^4)/(x^3 - 1)^2.
a(n) = 2*a(n-3) - a(n-6) for n>5.
a(n) = 27*n/(5 + 4*cos(2*n*Pi/3))^2.
If n mod 3 = 0, then n/3, else 3*n.
a(n) = lcm(numerator(n/3), denominator(n/3)). (End)
Sum_{k=1..n} a(k) ~ (19/18)*n^2. - Amiram Eldar, Oct 07 2023

Corrected and extended by Harvey P. Dale, May 16 2013

A102770 (p*q - 1)/2 where p and q are consecutive odd primes.

Original entry on oeis.org

7, 17, 38, 71, 110, 161, 218, 333, 449, 573, 758, 881, 1010, 1245, 1563, 1799, 2043, 2378, 2591, 2883, 3278, 3693, 4316, 4898, 5201, 5510, 5831, 6158, 7175, 8318, 8973, 9521, 10355, 11249, 11853, 12795, 13610, 14445, 15483, 16199, 17285, 18431, 19010

Offset: 1

W. Neville Holmes, Feb 10 2005

Primes in this sequence: 7, 17, 71, 449, 881, 2591, ... - Zak Seidov, Jan 14 2013

			a(1) = (3*5 - 1)/2 = 7.
a(2) = (5*7 - 1)/2 = 17.
a(3) = (7*11 - 1)/2 = 38.

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

Cf. A006094, A023515, A086870, A120876.

Table[(Prime[n] Prime[n + 1] - 1)/2, {n, 2, 50}] (* Alonso del Arte, Jan 14 2013 *)
(Times@@#-1)/2&/@Partition[Prime[Range[2,50]],2,1] (* Harvey P. Dale, Apr 10 2015 *)

a(n)=prime(n+1)*prime(n+2)\2 \\ Charles R Greathouse IV, Jan 14 2013

a(n) = (prime(n + 1)*prime(n + 2) - 1)/2.
a(n) ~ 0.5 n^2/log^2 n. - Charles R Greathouse IV, Jan 14 2013
a(n) = A023515(n+2)/2. - Jason Kimberley, Oct 23 2015

A096005 For k >= 1, let b(k) = ceiling( Sum_{i=1..k} 1/i ); a(n) = number of b(k) that are equal to n.

Original entry on oeis.org

0, 1, 2, 7, 20, 52, 144, 389, 1058, 2876, 7817, 21250, 57763, 157017, 426817, 1160207, 3153770, 8572836, 23303385, 63345169, 172190019, 468061001, 1272321714, 3458528995, 9401256521, 25555264765, 69466411833, 188829284972

Offset: 0

W. Neville Holmes, Jul 29 2004

nonn

			The ceilings of the first several partial sums of the reciprocal of the positive integers are 1 2 2 3 3 3 3 3 3 3 4 4 and the series is monotonically increasing, so a(0) = 0 (there being no zero), a(1) = 1 (there being but one 1) and a(3) = 7 (there being seven 3s).

Cf. A002387, A096143.

int A096005(int k)
{    if(k<3) return k;
    double sum = 0, n = 1; int ceiling = 2, cnt = 0;
    for(;;) {
        sum += 1/n++;
        if(sum < ceiling) { cnt++; continue; }
        if(ceiling++ == k) return cnt; else cnt = 1; }
} /* Oskar Wieland, May 01 2014 */

fh[0] = 0; fh[1] = 1; fh[k_] := Module[{tmp}, If[ Floor[tmp = Log[k + 1/2] + EulerGamma] == Floor[tmp + 1/(24k^2)], Floor[tmp], UNKNOWN]]; a[0] = 1; a[1] = 2; a[n_] := Module[{val}, val = Round[Exp[n - EulerGamma]]; If[fh[val] == n && fh[val - 1] == n - 1, val, UNKNOWN]]; Table[ a[n + 1] - a[n], {n, 0, 27}] (* Robert G. Wilson v, Aug 05 2004 *)

a(n+1)/a(n) approaches e = exp(1) = 2.71828...

First differences of A002387. - Vladeta Jovovic, Jul 30 2004

More terms from Robert G. Wilson v, Aug 05 2004

A096143 a(n) = ceiling(Sum_{i=1..n} 1/i).

Original entry on oeis.org

1, 2, 2, 3, 3, 3, 3, 3, 3, 3, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6

Offset: 1

W. Neville Holmes, Jul 24 2004

			a(4)=3 because the ceiling of 1 + 1/2 + 1/3 + 1/4 is 3.

Cf. A002387, A055980.

Ceiling[HarmonicNumber[Range[110]]] (* Harvey P. Dale, Aug 27 2014 *)

a(n) = ceil(sum(i=1, n, 1/i)) \\ Michel Marcus, Jul 11 2013

a(n) = ceiling(A001008(n)/A002805(n)). - Michel Marcus, Jul 11 2013

A074375 s(s+3)/2 where s is the sum of the prime factors of n (with repetition).

Original entry on oeis.org

0, 5, 9, 14, 20, 20, 35, 27, 27, 35, 77, 35, 104, 54, 44, 44, 170, 44, 209, 54, 65, 104, 299, 54, 65, 135, 54, 77, 464, 65, 527, 65, 119, 209, 90, 65, 740, 252, 152, 77, 902, 90, 989, 135, 77, 350, 1175, 77, 119, 90, 230, 170, 1484, 77, 152, 104, 275, 527, 1829, 90

Offset: 1

W. Neville Holmes, Aug 29 2002

			a(20) = 9(9+3)/2 = 54 because 9 = 2+2+5 and 20 = 2*2*5.

Neville Holmes, Integer Sequence Combinations

Applies A000096 to A001414. Cf. A074373, A074374.

spf[n_]:=Module[{c=Total[Times@@@FactorInteger[n]]},(c(c+3))/2]; Join[ {0}, Rest[Array[spf,60]]] (* Harvey P. Dale, Aug 16 2011 *)

A074373 Square of the sum of the prime factors of n (with repetition).

Original entry on oeis.org

0, 4, 9, 16, 25, 25, 49, 36, 36, 49, 121, 49, 169, 81, 64, 64, 289, 64, 361, 81, 100, 169, 529, 81, 100, 225, 81, 121, 841, 100, 961, 100, 196, 361, 144, 100, 1369, 441, 256, 121, 1681, 144, 1849, 225, 121, 625, 2209, 121, 196, 144, 400, 289, 2809, 121, 256

Offset: 1

W. Neville Holmes, Aug 28 2002

			a(12)=49 because 12=2*2*3, 2+2+3=7 and 7^2 = 49.

Harvey P. Dale, Table of n, a(n) for n = 1..1000
Neville Holmes, Integer Sequence Combinations

Equals A001414^2.

sspf[n_]:=Total[Flatten[Table[#[[1]],{#[[2]]}]&/@FactorInteger[ n]]]^2; Join[{0},Array[sspf,60,2]] (* Harvey P. Dale, Sep 24 2012 *)

cp's OEIS Frontend

User: W. Neville Holmes

A130699 Numbers n for which neither 2n-3 nor 2n+3 are primes.

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A130771 Numbers n not divisible by 3 for which neither 2n-3 nor 2n+3 are primes.

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A130723 Least common multiple of 3 and n^2+n+1.

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A130770 One third of the least common multiple of 3 and n^2+n+1.

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A130724 a(n) = lcm(n,3) / gcd(n,3).

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A102770 (p*q - 1)/2 where p and q are consecutive odd primes.

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A096005 For k >= 1, let b(k) = ceiling( Sum_{i=1..k} 1/i ); a(n) = number of b(k) that are equal to n.

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