Marian Kraus - cp's OEIS frontend

A287895 Differences of A287894.

4, 13, 12, 3, 61, 42, 50, 39, 43, 80, 65, 89, 180, 110, 146, 160, 156, 187, 88, 118, 258, 243, 244, 282, 287, 189, 343, 311, 203, 410, 219, 467, 473, 274, 384, 535, 406, 484, 54, 825, 310, 558, 551, 375, 886, 659, 539, 515, 294, 914, 764, 478, 774, 757, 823

Offset: 1

Marian Kraus, Jun 02 2017

The first two negative terms are a(99) = -239 and a(999) = -5966. - Giovanni Resta, Jun 07 2017

			Take the differences of 1,5,18,30,... => 4,13,12,...

Robert Israel, Table of n, a(n) for n = 1..1000

Cf. A287894.

ds:= n -> convert(convert(n,base,10),`+`):
f:= n -> add(ds(k^n),k=1..n):
A287894:= map(f, [$1..101]);
A287894[2..101] - A287894[1..100]; # Robert Israel, Jun 30 2017

s[n_] := Sum[Total@IntegerDigits[k^n], {k, n}]; Differences@ Array[s, 60] (* Giovanni Resta, Jun 07 2017 *)

A287894 Sum of the digit sums of the n-th powers of the first n positive integers.

Original entry on oeis.org

1, 5, 18, 30, 33, 94, 136, 186, 225, 268, 348, 413, 502, 682, 792, 938, 1098, 1254, 1441, 1529, 1647, 1905, 2148, 2392, 2674, 2961, 3150, 3493, 3804, 4007, 4417, 4636, 5103, 5576, 5850, 6234, 6769, 7175, 7659, 7713, 8538, 8848, 9406, 9957, 10332, 11218, 11877

Offset: 1

Marian Kraus, Jun 02 2017

The sequence is not increasing: a(100) < a(99). - Robert Israel, Jun 30 2017

			n=1: 1: 1;
n=2: 1,4: 1+4=5;
n=3: 1,8,27: 1+8+9=18;
n=4: 1,16,81,256: 1+7+9+13=30.

Robert Israel, Table of n, a(n) for n = 1..1000

Cf. A007953.

ds:= n -> convert(convert(n,base,10),`+`):
f:= n -> add(ds(k^n),k=1..n):
map(f, [$1..50]); # Robert Israel, Jun 30 2017

a[n_] := Sum[ Total@ IntegerDigits[k^n], {k, n}]; Array[a, 50] (* Giovanni Resta, Jun 07 2017 *)

a(n) = sum(k=1, n, sumdigits(k^n)); \\ Michel Marcus, Jun 06 2017

a(n) = Sum_{k=1..n} digsum(k^n), n >= 1.

A265207 Draw a square and follow these steps: Take a square and place at its edges isosceles right triangles with the edge as hypotenuse. Draw a square at every new edge of the triangles. Repeat for all the new squares of the same size. New figures are only placed on empty space. The structure is symmetric about the first square. The sequence gives the numbers of squares of equal size in successive rings around the center.

Original entry on oeis.org

1, 8, 20, 36, 60, 92, 140, 204, 300, 428, 620, 876, 1260, 1772, 2540, 3564, 5100, 7148, 10220, 14316, 20460, 28652, 40940, 57324, 81900, 114668, 163820, 229356, 327660, 458732, 655340, 917484, 1310700, 1834988, 2621420, 3669996, 5242860, 7340012, 10485740, 14680044, 20971500, 29360108, 41943020, 58720236

Offset: 1

Marian Kraus, Dec 04 2015

			By recursion:
a(3)=2*a(1)+20=2*8+20=36
a(4)=2*a(2)+20=2*20+20=60
By function:
a(3)=4*sum_{k=1}^{[(3+1)/2]}(2^k)+6*sum_{k=1}^{[3/2]}(2^k)
=4*sum_{k=1}^{[2]}(2^k)+6*sum_{k=1}^{[1.5]}(2^k)
=4*sum_{k=1}^{2}(2^k)+6*sum_{k=1}^{1}(2^k)
=4*(2^1+2^2)+6*(2^1)
=4*(2+4)+6*(2)=24+12=36
a(4)=4*sum_{k=1}^{[(4+1)/2]}(2^k)+6*sum_{k=1}^{[4/2]}(2^k)
=4*sum_{k=1}^{[2.5]}(2^k)+6*sum_{k=1}^{[2]}(2^k)
=4*sum_{k=1}^{2}(2^k)+6*sum_{k=1}^{2}(2^k)
=4*(2^1+2^2)+6*(2^1+2^2)
=4*(2+4)+6*(2+4)=24+36=60

Marian Kraus, Illustration for a(4)

For the differences (a(n)-a(n-1))/4, n>2, see A163978.

Cf. A029744, A063759, A164090.

rm(a)
a <- vector() powerof2 <- vector()
x <- 300
n <- x/2
for (i in 1:x){
   powerof2[i] <- 2^(i-1)}
powerof2 for (i in 1:n){
   a[2*i]   <- 8*(sum(powerof2[1:i]))+12*(sum(powerof2[1:i]))}
for (i in 1:(n+1)){
   a[2*i+1] <- 8*(sum(powerof2[1:(i+1)]))+12*(sum(powerof2[1:i]))}
a[1]<-8
a

Conjectured recurrence:
a(0)=1,
a(1)=8,
a(2)=20, and thereafter
a(n)=2*a(n-2)+20.
Conjectured formula: ("[]" is the floor function)
a(n)=4*sum_{k=1}^{[(n+1)/2]}(2^k)+6*sum_{k=1}^{[n/2]}(2^k).
Conjectures from Colin Barker, Dec 07 2015: (Start)
a(n) = (-20+2^(1/2*(-1+n))*(10-10*(-1)^n+7*sqrt(2)+7*(-1)^n*sqrt(2))) for n>1.
a(n) = 5*2^(n/2+1/2)-5*(-1)^n*2^(n/2+1/2)+7*2^(n/2)+7*(-1)^n*2^(n/2)-20 for n>1.
a(n) = a(n-1)+2*a(n-2)-2*a(n-3) for n>4.
G.f.: x*(1+7*x+10*x^2+2*x^3) / ((1-x)*(1-2*x^2)).
(End)

A256592 Let p = prime(n); a(n) = number of pairs (x,i) with i >= 2 and 2 <= x <= p-i such that x(x+1)(x+2)...(x+i-1) == 1 mod p.

Original entry on oeis.org

0, 0, 1, 2, 6, 3, 8, 7, 13, 15, 13, 11, 13, 22, 18, 25, 36, 31, 34, 53, 42, 38, 38, 40, 55, 47, 41, 37, 77, 59, 62, 67, 66, 63, 55, 84, 74, 78, 90, 74, 90, 92, 85, 108, 100, 117, 98, 104, 136, 114, 118, 118, 141, 112, 118, 115, 122, 138, 132, 129, 115, 152

Offset: 1

Marian Kraus, Apr 03 2015

nonn

			prime(1)=2: There is no such product
=> a(1)=0;
prime(2)=3: There is no such product
=> a(2)=0;
prime(3)=5: 2*3=6==1 mod 5
=> i=1, x=2; a(3)=1;
prime(4)=7: 4*5*6==1 mod 7; 2*3*4*5==1 mod 7
=> a(3)=2;
prime(5)=11: 3*4==1 mod 11; 7*8==1 mod 11; 5*6*7==1 mod 11; 3*4*5*6*7==1 mod 11; 6*7*8*9*10==1 mod 11; 2*3*4*5*6*7*8*9==1 mod 11
=> x in {3,7,5,3,6,2}
=> a(5)=6.

Cf. A256567, A256580.

f[n_] := Block[{r = Range[2, Prime[n] - 1]}, Sum[Length@ Select[Times @@@ Partition[r, k, 1], Mod[#, Prime@ n] == 1 &], {k, 2, Prime@ n}]]; Array[f, 72] (* Michael De Vlieger, Apr 03 2015 *)

library(numbers)
p <- vector()
n <- vector()
NumTup <- vector()
p <- Primes(m)
n <- length(p)
m <- 17 #all primes will be checked up to this number
Piprod <- matrix(0,m,m) #Matrix with zeros
#loop: every ordered combination of products
for (i in 2:m)
for (j in 2:m)
  Piprod[j,i] <- ifelse(i

A256580 Number of quadruples (x, x+1, x+2, x+3) with 1 < x < p-3 of consecutive integers whose product is 1 mod p.

Original entry on oeis.org

0, 0, 0, 1, 0, 0, 2, 0, 3, 0, 2, 0, 2, 0, 2, 0, 0, 0, 0, 0, 2, 0, 0, 2, 2, 0, 2, 0, 0, 0, 4, 0, 4, 0, 0, 4, 0, 0, 2, 0, 0, 0, 0, 0, 0, 2, 0, 2, 0, 0, 4, 4, 2, 0, 2, 0, 0, 2, 0, 4, 0, 0, 0, 2, 2, 0, 0, 0, 0, 0, 2, 4, 2, 0, 0, 2, 0, 0, 0, 2, 0, 0, 4, 2, 2, 0, 4, 0, 0, 0, 0, 2, 4, 0, 0, 2, 0, 2, 0, 0, 0, 0, 0, 0, 0, 2, 0, 2, 0, 2, 2, 0, 0, 0, 0, 0, 0, 2, 0, 0, 0, 4, 0, 0, 0, 0, 0, 2, 2, 0, 0, 4, 4, 0, 2, 2, 0, 0, 0, 0, 0, 0, 0, 0, 0, 2, 0, 2, 0, 0, 0, 2, 0, 2, 0, 4, 0, 2, 2, 0, 0, 4, 4, 0, 4, 2, 0, 0

Offset: 1

Marian Kraus, Apr 02 2015

nonn

If "quadruples" is changed to "pairs" we get A086937 (for the counts) and A038872 (for the primes for which the count is nonzero).

			p=7, x_1=2, 2*3*4*5 == 1 (mod 7), T={2}, |T|=1;
p=17, x_1=2, 2*3*4*5 == 1 (mod 17), x_2=12, 12*13*14*15 == 1 (mod 17), T={2,12}, |T|=2;
p=23, x_1=5, 5*6*7*8 == 1 (mod 23), x_2=15, 15*16*17*18 == 1 (mod 23), x_3=19, 19*20*21*22 == 1 (mod 23), T={5,15,19}, |T|=3.

Cf. A256567, A256572, A038872, A086937.

library(numbers);IP <- vector();t <- vector();S <- vector();IP <- c(Primes(1000));for (j in 1:(length(IP))){for (i in 2:(IP[j]-4)){t[i-1] <-as.vector(mod((i*(i+1)*(i+2)*(i+3)),IP[j]));Z[j] <- sum(which(t==1));S[j] <- length(which(t==1))}};S

|T| where T = {x|x*(x+1)*(x+2)*(x+3) == 1 mod p, p is prime, 1 < x < p-3}.

A256567 Primes p with the property that there are three consecutive integers (x,x+1,x+2) with 1 < x <= p-3 whose product is 1 modulo p.

Original entry on oeis.org

7, 11, 17, 19, 23, 37, 43, 53, 59, 61, 67, 79, 83, 89, 97, 101, 103, 107, 109, 113, 137, 149, 157, 167, 173, 181, 191, 199, 211, 223, 227, 229, 241, 251, 263, 271, 281, 283, 293, 307, 313, 317, 337, 347, 359, 367, 373, 379, 383, 389, 401, 419, 421, 431, 433, 449

Offset: 1

Marian Kraus, Apr 02 2015

nonn

There may be one or more such triples, but 23 is the only prime up to 100000 having precisely two such triples. For the number of triples for each prime, see A256572.

Together with 5, supersequence of A191065. - Arkadiusz Wesolowski, Nov 24 2021

			For p=7: 4*5*6=120==1 (mod 7), so 7 is a term.
For p=11: 5*6*7=210==1 (mod 11), so 11 is a term.
For p=17: 4*5*6=120==1 (mod 17), so 17 is a term.
13 is not a term because there is no such triple with product ==  1 (mod 13).

Marian Kraus, Table of n, a(n) for n = 1..6390

Cf. A191065, A256572, A254678, A256580.

isok(p) = {if (isprime(p), for (x=1, p-3, if (Mod(x*(x+1)*(x+2), p) == 1, return (1));););} \\ Michel Marcus, Oct 05 2021

library(numbers)
IP <- vector()
t <- vector()
S <- vector()
IP <- c(Primes(1000)) # Build a vector of all primes < 1000.
for (j in 1:(length(IP))){
   for (i in 3:(IP[j]-2))
      t[i-1] <- as.vector(mod(((i-1)*i*(i+1)),IP[j]))
   S[j] <- length(which(t==1))
}
IP[S!=0]
#The loop checks for every triple for every prime, what it is modulo that prime. "IP[S!=0]" lists the primes that have at least one triple. For all p<10000 it takes a few minutes. For all p<100000 a few hours.

A256572 Number of triples (x,x+1,x+2) with 1 < x <= p-3 of consecutive integers less than p whose product is 1 modulo p, where p = prime(n).

Original entry on oeis.org

0, 0, 0, 1, 1, 0, 1, 1, 2, 0, 0, 1, 0, 1, 0, 1, 3, 1, 1, 0, 0, 1, 1, 1, 1, 3, 1, 1, 1, 1, 0, 0, 1, 0, 1, 0, 1, 0, 3, 3, 0, 1, 1, 0, 0, 1, 3, 3, 1, 1, 0, 0, 1, 1, 0, 1, 0, 3, 0, 1, 1, 1, 3, 0, 1, 3, 0, 1, 3, 0, 0, 1, 1, 1, 1, 1, 1, 0, 1, 0, 1, 1, 1, 1, 0, 0, 3, 1, 0, 3, 1, 1, 0, 0, 0, 1, 0, 1, 1, 0, 0, 1, 1, 1, 1, 0, 0, 3, 3, 0, 3, 1, 1, 1, 1, 1, 1, 0, 0, 1, 1, 0, 1, 0, 3, 1, 1, 3, 1, 1, 0, 1, 1, 1, 0, 1, 1, 1, 1, 3, 0, 3, 0, 1, 3, 1, 3, 0, 0, 0, 3, 1, 3, 0, 1, 1, 1, 0, 1, 1, 0, 1, 0, 1, 1, 1, 3, 3

Offset: 1

Marian Kraus, Apr 02 2015

nonn

Is 23 the only prime with two triples?

			For p=7: 4*5*6==1 (mod 7); T={4}, |T|=1.
For p=23: 2*3*4==1 (mod 23) and 9*10*11==1 (mod 23); T={2,9}, |T|=2.
For p=59: 3*4*5==1 (mod 59), 12*13*14==1 (mod 59), and 41*42*43==1 (mod 59); T={3,12,41}, |T|=3.

Marian Kraus, Table of n, a(n) for n = 1..9592

Cf. A256567.

a(n) = {my(p = prime(n)); sum(x=2, p-3, (x*(x+1)*(x+2)) % p == 1);} \\ Michel Marcus, Apr 03 2015

library(numbers);IP <- vector();t <- vector();S <- vector();IP <- c(Primes(1000));LIP <- length(IP);for (j in 1:LIP){for (i in (3:(IP[j]-2))){t[i-1] <- as.vector(mod(((i-1)*i*(i+1)),IP[j]))};S[j] <- length(which(t==1))};S
#Needs a lot of memory. For Primes(100000), this takes a few hours.

|T| where T={x|x*(x+1)*(x+2)==1 (mod p), p is prime, 1

A254678 Primes p with the property that there are four consecutive integers less than p whose product is 1 mod p.

Original entry on oeis.org

7, 17, 23, 31, 41, 47, 73, 89, 97, 103, 127, 137, 151, 167, 199, 223, 233, 239, 241, 257, 271, 281, 311, 313, 353, 359, 367, 383, 409, 431, 433, 439, 449, 479, 487, 503, 521, 577, 593, 601, 607, 647, 673, 719, 727, 743, 751, 761, 769, 839, 857, 881, 887, 911, 929, 937, 953, 967, 977, 983

Offset: 1

Marian Kraus, Apr 02 2015

nonn

			p=7: 2*3*4*5=120 == 1 mod 7;
p=17: 2*3*4*5=120 == 1 mod 17 AND 12*13*14*15=32760 == 1 mod 17; for p=13: no triple == 1 mod 13;
p=23: 5*6*7*8 == 1 mod 23 AND 15*16*17*18== 1 mod 23 AND 19*20*21*22 == 1 mod 23; and so on. For the number of quadruples for a prime, see A256580.

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

Cf. A256567, A256580.

fsiQ[n_]:=AnyTrue[Times@@@Partition[Range[n-1],4,1],Mod[#,n]==1&]; Select[ Prime[Range[200]],fsiQ] (* Requires Mathematica version 10 or later *) (* Harvey P. Dale, Jul 02 2019 *)

lista(nn) = forprime(p=2, nn, if (sum(x=1, p-4, ((x*(x+1)*(x+2)*(x+3)) % p) == 1) > 0, print1(p, ", "))); \\ Michel Marcus, Apr 03 2015

library(numbers)
IP <- vector()
t <- vector()
S <- vector()
IP <- c(Primes(1000))
for (j in 1:(length(IP))){
   for (i in 2:(IP[j]-4)){
       t[i-1] <- as.vector(mod((i*(i+1)*(i+2)*(i+3)),IP[j]))
       Z[j] <- sum(which(t==1))
       S[j] <- length(which(t==1))
   }
}
IP[S!=0]
#Carefully increase Primes(1000). It takes several hours for 100000.

x*(x+1)*(x+2)*(x+3) == 1 mod p, p is prime, 1 <= x <= p-4.

A113397 What are the values of k in the term Prime(n+1)^2-Prime(n)^2 = a+k*(Prime(n+1)) if "a" is element of {0,1,...,Prime(n+1)-1}.

Original entry on oeis.org

1, 3, 3, 6, 3, 7, 3, 7, 10, 3, 11, 7, 3, 7, 11, 11, 3, 11, 7, 3, 11, 7, 11, 15, 7, 3, 7, 3, 7, 26, 7, 11, 3, 19, 3, 11, 11, 7, 11, 11, 3, 19, 3, 7, 3, 23, 23, 7, 3, 7, 11, 3, 19

Offset: 1

Marian Kraus, Oct 26 2005

nonn

			3^2-2^2=2+1*3; 5^2-3^2=1+3*5; 7^2-5^2=3+3*7; 11^2-7^2=6+6*11; 13^2-11^2=9+3*13; etc.

Cf. A113396.

Prime(n+1)^2-Prime(n)^2 = a+k*(Prime(n+1)) "a" element of {0, 1, ..., Prime(n+1)-1}

A113398 How many times did the value of k appear at the previous primes including the n-th one? (Prime(n+1))^2-(Prime(n))^2 = a+k*(Prime(n+1)) if a is element of {0,1,...,Prime(n+1)-1}.

Original entry on oeis.org

1, 1, 2, 1, 3, 1, 4, 2, 1, 5, 1, 3, 6, 4, 2, 3, 7, 4, 5, 8, 5, 6, 6, 1, 7, 9, 8, 10, 9, 1, 10, 7, 11, 1, 12, 8, 9, 11, 10, 11, 13, 2, 14, 12, 15, 1, 2, 13, 16, 14, 12, 17, 3

Offset: 1

Marian Kraus, Oct 26 2005

nonn

			3^2-2^2 = 2+1*3 (first "1"); 5^2-3^2 = 1+3*5 (first "3"); 7^2-5^2 = 3+3*7 (2nd "3");
11^2-7^2 = 6+6*11 (first "6"); 13^2-11^2 = 9+3*13 (3rd "3"); etc.

Cf. A113396, A113397.

cp's OEIS Frontend

User: Marian Kraus

A287895 Differences of A287894.

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A287894 Sum of the digit sums of the n-th powers of the first n positive integers.

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A256592 Let p = prime(n); a(n) = number of pairs (x,i) with i >= 2 and 2 <= x <= p-i such that x*(x+1)*(x+2)*...*(x+i-1) == 1 mod p.

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Mathematica

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A256580 Number of quadruples (x, x+1, x+2, x+3) with 1 < x < p-3 of consecutive integers whose product is 1 mod p.

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A256567 Primes p with the property that there are three consecutive integers (x,x+1,x+2) with 1 < x <= p-3 whose product is 1 modulo p.

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A256572 Number of triples (x,x+1,x+2) with 1 < x <= p-3 of consecutive integers less than p whose product is 1 modulo p, where p = prime(n).

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A254678 Primes p with the property that there are four consecutive integers less than p whose product is 1 mod p.

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A256592 Let p = prime(n); a(n) = number of pairs (x,i) with i >= 2 and 2 <= x <= p-i such that x(x+1)(x+2)...(x+i-1) == 1 mod p.